Basic data structures

Abstract

thanks for the help by simonmysun for the Collection in learning and practicing of data structure

the following comes the solutions and summary of data structure:

Use arrays to implement linked lists.

there are various ways to implement a linked list with using arrays. here i wrote tow different ways:

1. using Node structure and pointer(Singly Linked List).
each Node contains a data type and a pointer to the next Node:
struct node {
int data;
struct node *next;
};


besides we wil use a function to conected each 2 "adjacent" Nodes. It has tow Parameters:
array pointer and size of array:

#include <iostream>
#include <string>
#include <stdlib.h>
using namespace std;

struct node {
int data;
struct node *next;
};

node * createnode(int val) {
node * newnode = (node*)malloc(sizeof(node));
if (newnode == NULL) {
return NULL;
} else {
newnode->data = val;
newnode->next = NULL;
return newnode;
}
}

int main() {
int arr[100];
int n;
cin >> n;
for (int k = 0; k < n; k++) {
cin >> arr[k];
}
node * tail = NULL;
for (int i = 0; i < n; i++) {
if (i == 0) {
} else {
node * newnode = createnode(arr[i]);
tail->next = newnode;
tail = tail->next;
}
}
while (element != NULL) {
cout << element->data << " ";
element = element->next;
}
cout << endl;
}


pointer function node * createnode(int val) return a new node with a empty pointer to NULL
in the main func we use loop to create new node meanwhile we keep the head value and tail value to the next
for operation of the linked list we just need head as input.
for remove:

void DeleteByVal(node *head, int val)
{
{
return;
}

//find target node and its precursor
while(cur)
{
if (cur->data == val)
break;
else {
cur = cur->next;
pre = pre->next;
}
}

//delete target node
pre->next = cur->next;
free(cur);
}


for empty the list:

void Free(node *head)
{
{
free(temp);
}
}

1. using array
#include <iostream>
using namespace std;
struct node {
int data;
int next;
};
node A[10];
int main() {
int arr[10];
for (int i = 0; i < 10; i++) {
cin >> arr[i];
A[i].data = arr[i];
if (i != 9) {
A[i].next = A[i+1].data;
} else {
A[i].next = A[0].data;
}
}
for (int j = 0; j < 10; j++) {
cout << A[j].data << " ";
}
}



we can write directly with out doubt:

node* Reverse (node* head) {
else {
*pre = NULL,
*next = NULL;
while (cur != NULL) {
next = cur->next;
cur->next = pre;
pre = cur;
cur = next;
}
}
}


Find n th node from the end of a given linked list.

node * find_r_n_node(node * head， int n) {
for (int i = 0; i < n-1; i++) {
fast = fast->next;
}
while (fast != NULL && fast->next != NULL) {
fast = fast->next;
slow = slow->next;
}
return slow;
}


Find the middle node of a given linked list.

node * slow = head;
while(fast != NULL && fast->next != NULL) {
slow = slow->next;
fast = fast->next->next;
}
return slow;


• If you get stucked on this problem, you may also try the following problems first.
• Find and delete a specified node in a given linked list.
void find_element(node* head, int* target) {
while (element != target) {
element = element->next;
}
element->pre->next = element->next;
element->next->pre = element->pre;
}


• Swap two nodes on a given linked list.
for a linked list with size n, if we wanna swap tow nodes (i and j for 0 =< i, j < n)
named as node_i and node _j.
with find_element we can easily find the tow previous and next nodes of *i* and *j* nodes. just named
as prenode_i, prenode_j and nextnode_i, nextnode_j. then we can reset prenode_i->next = node_j, node_j->next = nextnode_i
prenode_j = node_i, node_i->next = nextnode_j .
void swap_node(node* a, node* b) {
node * temp_a = new node;
node * temp_b = new node;
*temp_a = *a;
*temp_b = *b;
*a = *b;
a->pre = temp_a->pre;
a->next = temp_a->next;
*b = *temp_a;
b->pre = temp_b->pre;
b->next = temp_b->next;
delete temp_a;
delete temp_b;
}


for example 10 nodes linked list and swap 3 and 6, following is the result:

1 2 3 4 5 6 7 8 9 10

1 2 6 4 5 3 7 8 9 10

• Implement bubble sort on linked lists.

classic bubble sort for array in c++ WiKi:

#include <iostream>
#include <algorithm>
using namespace std;
template<typename T> //整數或浮點數皆可使用,若要使用物件(class)時必須設定大於(>)的運算子功能
void bubble_sort(T arr[], int len) {
int i, j;
for (i = 0; i < len - 1; i++)
for (j = 0; j < len - 1 - i; j++)
if (arr[j] > arr[j + 1])
swap(arr[j], arr[j + 1]);
}
int main() {
int arr[] = { 61, 17, 29, 22, 34, 60, 72, 21, 50, 1, 62 };
int len = (int) sizeof(arr) / sizeof(*arr);
bubble_sort(arr, len);
for (int i = 0; i < len; i++)
cout << arr[i] << ' ';
cout << endl;
float arrf[] = { 17.5, 19.1, 0.6, 1.9, 10.5, 12.4, 3.8, 19.7, 1.5, 25.4, 28.6, 4.4, 23.8, 5.4 };
len = (int) sizeof(arrf) / sizeof(*arrf);
bubble_sort(arrf, len);
for (int i = 0; i < len; i++)
cout << arrf[i] << ' ';
return 0;
}


base on this code we can implement bubble sort for Linked List.
first build a module for swap two adjacent elements of the linked list. this function returns the element which was
beyond the another and after swapping goes to be behind it. and i is the swap position for i and i+1, n is the

void bubble_sort(node * head) {
do {
while (element != find_r_n_node(head, i)) {
if element->data > element->next->data;
swap_node(element, element->next);
element = element->next;
i++;
}
}

• Given an ordered linked list, insert a new number without destroying its order.
void insert_ordered_ll(node ** head, node * new_e) {
return;
}
for (node * element = *head; element != NULL; element = element->next) {
if (element->next == NULL) {
if (new->data >= element->data) {
element->next = new;
return;
}
}
if (new_e->data >= element->data && new_e->data <= element->next->data) {
element->next = new_e;
new_e->next = element->next;
return;
}
}
}


following is the result of an example:

1 2 9 10 11 -> 1 2 6 9 10 11

• Implement insertion sort on linked lists.
fisrt of all we build a double linked list. struct node contains 3 elements: value of single node, pointer to
next node, pointer to previous node.
struct node {
int data;
struct node * pre;
struct node * next;
};


with function create_node(int val) to create every node in loop.

node* create_node(int val) {
node * newnode = (node*)malloc(sizeof(node));
if (newnode == NULL) {
cout << "out of memory !" << endl;
return NULL;
} else {
newnode->data = val;
newnode->next = NULL;
newnode->pre = NULL;
return newnode;
}
}


node * insert_double_linked_list(node * head, node * tail, int val) to connect current tail with new node and
return new tail.

node * insert_double_linked_list(node * head, node * tail, int val) {                                                             node *newnode = create_node(val);
if (!tail) {
tail = newnode;
}
else {
tail->next = newnode;
newnode->pre = tail;
tail = newnode;
}
return tail;
}


based on insert sort for array WiKi:

void insertion_sort(int arr[],int len){
for(int i=1;i<len;i++){
int key=arr[i];
int j;
for(j=i-1;j>=0 && key<arr[j];j--)
arr[j+1]=arr[j];
arr[j+1]=key;
}
}


we can easily implement insert sort for linked list:

void insertion_sort(node * head) {
for (node * m_node = head->next; m_node != NULL; m_node = m_node->next) {
int key = m_node->data;
node * s_node;
for (s_node = m_node->pre; s_node != NULL && key < s_node->data; s_node = s_node->pre) {
s_node->next->data = s_node->data;
}
if (s_node) s_node->next->data = key;
}
}


following is the left code:

int main() {
int arr[100];
int n;
cin >> n;
for (int i = 0; i < n; i++) cin >> arr[i];
node * tail;
for (int i = 0; i < n; i++) {
if (i == 0) {
}
for (node * element = head; element != NULL; element = element->next) cout << element->data << " ";
cout << endl;
for (node * element = tail; element != NULL; element = element->pre) cout << element->data << " ";
cout << endl;
for (node * element = head; element != NULL; element = element->next) cout << element->data << " ";
cout << endl;
}


here is the result:

input:
10

10 5 2 7 6 8 3 9 1 4

output:

10 5 2 7 6 8 3 9 1 4

4 1 9 3 8 6 7 2 5 10

1 2 3 4 5 6 7 8 9 10

• Given a linked list, divide them into two even halves.

• Given two ordered linked lists, merge them into one ordered linked list.

• Implement merge sort on linked lists.

these three we can solve in a row.

if we wanna implement merge sort, first we need divide linked list into two halves
hence we use double pointer node** head_1 and node** head_1 to receive the two head of lists which was spilted

void divide_node(node * head, node** head_1, node** head_2) {
while(fast) {
if (fast->next) fast = fast->next->next;
else break;
slow = slow->next;
}
slow->next = NULL;
}


then we build a function node * merge_node(node* head_1, node* head_2) which based on recursion. input two

node * merge_node(node * head_1, node * head_2) {
}
else {
}
}


finally we implement a merge sort , but we must consider that using recursion to spilt node as single and then merge

void merge_sort(node** head) {
cout << "OK" << endl;
}

• Given a linked list, divide them into two halves(might not be even) and meanwhile let each number in the first half be greater than all numbers in the second half.

• Implement quick sort on linked lists.

we can solve these two questions as a row.

first we need a swap fucction to swap two nodes.

void swap_node(node* a, node* b) {
node * temp_a = new node;
node * temp_b = new node;
*temp_a = *a;
*temp_b = *b;
*a = *b;
a->pre = temp_a->pre;
a->next = temp_a->next;
*b = *temp_a;
b->pre = temp_b->pre;
b->next = temp_b->next;
delete temp_a;
delete temp_b;
}


then we build a function partion to put the numbers lower than pivot to the left and the numbers greater than pivot
to the right.

node* get_partion(node* head, node * end) {
node* q = p->next;

while(q != end) {
if(q->data < key) {
p = p->next;
swap_node(p, q);
}
q = q->next;
}
return p;
}


the main function for quick sort. Recur for the list before pivot and after the pivot element.

void quick_sort(node * head, node * end) {
node * partion = get_partion(head, end);
quick_sort(partion->next, end);
}
}


Given a linked list(assume it is), tell whether there is a loop and find the entry of it.

#include <iostream>
using namespace std;

int main() {
while (fast) {
if (slow == fast) break;
slow = slow->next;
fast = fast->next->next;
}
node * p = slow;
while(new_element == p) {
new_element = new_element->next;
p = p->next;
}
node * entry = p;
}


Given two linked list, tell whether and where they intersect each other. What if there can be loops?

find_intersection_point(node * head1, node * head2) {
for (node * element1 = head1; element1 != NULL; element1 = element1->next)
for(node * element2 = head2; element2 != NULL; elemnt2 = element2->next)
if (element1 == element2) return element1;
}

node * find_intersection_point_loop(node * head1, node * head2) {
node * tail;
while(fast != slow) {
if (fast->next || fast->next->next) {
tail = fast;
}
fast = fast->next->next;
slow = slow->next;
}
node * p = slow;
while(new_element == p) {
new_element = new_element->next;
p = p->next;
}
tail->next = NULL;
return p;
}


Stack

implement a stack

Use array to implement stacks.

• Push a node into a given stack.

• Pop a node from a given stack.

• Peak the top node of a given stack.

• Empty a given stack.

#include <iostream>
#include <bits/stdc++.h>
using namespace std;

#define n 1000
class Stack {
int top;
public:
int a[n];
Stack() { top = -1;}
bool push(int x);
int pop();
bool empty();
};

bool Stack::push(int x) {
if (top >= n-1) {
cout << " Stack overflow" << endl;
return false;
} else {
a[++top] = x;
return true;
}
}

int Stack::pop() {
if (top < 0) {
cout << " Stack underflow" << endl;
return 0;
} else {
int x = a[top--];
return x;
}
}

void Stack::empty() {
top = -1;
}

int Stack::peak_top() {
if (top >= 0) return a[top];
}


Use pointers or references to implement stacks. Including the operations above.

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <iostream>
using namespace std;

struct StackNode {
int data;
struct StackNode* next;
};

void * push(StackNode** top, int value) {
StackNode * new_node = (StackNode*) malloc(sizeof(StackNode));
new_node->data = value;
new_node->next = *top;
*top = new_node;
cout << new_node->data << " pushed to stack" << endl;
}

int pop(StackNode** top) {
if(!*top) return -1;
StackNode* temp = *top;
*top = (*top)->next;
int popped = temp->data;
free(temp);
return popped;
}

int peak(StackNode* top) {
if (!top) return -1;
}

int main() {
StackNode* top = NULL;
push(&top, 1);
push(&top, 2);
push(&top, 3);
cout << "poped is " << pop(&top) << endl;
cout << "top is " << peak(top) << endl;
}


Given a sequence of push operations and a sequence of pop operations, tell whether it can be valid.

see the first quiz...

Implement a queue supporting push(). pop() and getMin().

#include <iostream>
#include <bits/stdc++.h>
#include <limits>
#include <cstring>
using namespace std;

#define n 1000
class Stack {
int top;
int* *MAX;
int* *MIN;
int m_top;
int min_top;
int a[n];
int MIN_INT = INT_MIN;
int MAX_INT = INT_MAX;
public:
Stack() {
top = -1;
m_top = 0;
min_top = 0;
MAX = new int* [n];
MIN = new int* [n];
*MAX = &MIN_INT;
*MIN = &MAX_INT;
}
bool push(int x);
int pop();
int peak_top();
void empty();
};

bool Stack::push(int x) {
if (top >= n-1) {
cout << " Stack overflow" << endl;
return false;
} else {
a[++top] = x;
if (x >= *MAX[m_top]) MAX[++m_top] = &a[top];
if (x <= *MIN[min_top]) MIN[++min_top] = &a[top];
cout << "after push, MAX = " << *MAX[m_top] << " MIN = " << *MIN[min_top] << endl;
return true;
}
}

int Stack::pop() {
if (top < 0) {
cout << " Stack underflow" << endl;
return 0;
} else {
if (&a[top] == MAX[m_top]) --m_top;
if (&a[top] == MIN[min_top]) --min_top;
int x = a[top--];
cout << "after pop, MAX = " << *MAX[m_top] << " MIN = " << *MIN[min_top] << endl;
return x;
}
}

void Stack::empty() {
top = -1;
}

int Stack::peak_top() {
if (top >= 0) return a[top];
}

int main() {
Stack stack1;
stack1.push(2);
stack1.push(5);
stack1.push(3);
stack1.push(6);
stack1.push(1);
stack1.push(9);
cout << stack1.peak_top() << endl;
cout << stack1.pop() << endl;
stack1.empty();
}


after push, MAX = 2 MIN = 2

after push, MAX = 5 MIN = 2

after push, MAX = 5 MIN = 2

after push, MAX = 6 MIN = 2

after push, MAX = 6 MIN = 1

after push, MAX = 9 MIN = 1

9

after pop, MAX = 6 MIN = 1

9

undo and redo

Implement undo/redo functionality(or back/forward navigation in explorer).

For some softwares, they generally offer the Redo and Undo Operations, like Action panel in Paint.NET, or UNDO/REDO in
Word , normally we just press CTRL-Z to implement it.
to implement the funtionality , directly thinking is, that we put the Command as a class into a Stack, which we can imagine it as
a class.
abstract the Command or Operation as a class, which can use different Request parameter to initialize the class, in other word Queue processing for Commands, recording the requests, execute Undo/Redo Operations., that's called Command Pattern. the
most advantages its is that, Separating the call and implement.

The standard is to keep the Command objects in a stack to support multi level undo. In order to support redo, a second stack keeps all the commands you've Undone. So when you pop the undo stack to undo a command, you push the same command you popped into the redo stack. You do the same thing in reverse when you redo a command. You pop the redo stack and push the popped command back into the undo stack.
here is a small example for implementing of undo/redo functionality:

first we build a command class ICoomand and its two subclass as Operator 1 and 2, here we used virtual function (see more about virtual function). Then we built a class AssWeCan for implementing of stack, whitch
contains undo-stack and redo-stack and a function void RBQ(command) to store the commands. class FAQ contains
end two outputs which can be called by two subclass of ICommand. otherwise, the two pointer functions (PopUndoCommand and
PopRedoCommand) were used to implement undo/redo, return the one which was poped from the top of stack and into anonther
stack.
for int main() function, we use while loop to implement multi commands and multi undo/redo.
see more deep dark fantasy.

here is the code by 'c++' :

#include <iostream>
#include <stack>
#include <stdio.h>
using namespace std;

class ICommand {
public:
virtual void Execute() = 0;
};

class AssWeCan {
public:
stack<ICommand *>undolist;
stack<ICommand *>redolist;
void RBQ(ICommand *command) {
if (command) {
undolist.push(command);
command->Execute();
}
}
};

class FAQ {
public:
void deedar() {
cout << "deep \u2642 dark \u2642 fantasy \u2642." << endl;
}
void perform_artist() {
cout << "do u like \u2642 what \u2642 u \u2642 see." << endl;
}
};

class DeedarCommand : public ICommand {
private:
FAQ *_fuck_u;
public:
DeedarCommand(FAQ *fuck_u) {
_fuck_u = fuck_u;
}
void Execute() {
_fuck_u->deedar();
}
};

class PerformCommand : public ICommand {
private:
FAQ *_fuck_u;
public:
PerformCommand(FAQ *fuck_u) {
_fuck_u = fuck_u;
}
void Execute() {
_fuck_u->perform_artist();
}
};

ICommand * PopUndoCommand(AssWeCan* asswecan) {
ICommand *pcommand = NULL;
if (!asswecan->undolist.empty()) {
pcommand = asswecan->undolist.top();
asswecan->undolist.pop();
}
return pcommand;
}

ICommand * PopRedoCommand(AssWeCan* asswecan) {
ICommand * pcommand = NULL;
if (!asswecan->redolist.empty()) {
pcommand = asswecan->redolist.top();
asswecan->redolist.pop();
}
return pcommand;
}

int main() {
stack<ICommand *>undolist;
stack<ICommand *>redolist;
FAQ* fuck_u = new FAQ;
ICommand* ddf = new DeedarCommand(fuck_u);
ICommand* pa = new PerformCommand(fuck_u);
AssWeCan* asswecan = new AssWeCan;
char ch;
cout << "Enter ur favourite w\u2642rd a)deedar b)perform artist: " << endl;
cin >> ch;
while (ch != 'q') {
if (ch == 'a') asswecan->RBQ(ddf);
else asswecan->RBQ(pa);
cout << "Enter ur favourito w\u2642rd a)deedar b)perform aritist (press q to quit): " << endl;
cin >> ch;
}
cout << asswecan->undolist.top() << endl;
asswecan->undolist.top()->Execute();
cout << "Press (u) to undo and (r) to redo( (q) for quit): " << endl;
char ch1;
cin >> ch1;
while(ch1 != 'q') {
if (ch1 == 'u' ) {
ICommand * pcommand = PopUndoCommand(asswecan);
cout << asswecan->undolist.top() <<endl;
asswecan->redolist.push(pcommand);
asswecan->undolist.top()->Execute();
}
else {
ICommand * pcommand = PopRedoCommand(asswecan);
asswecan->undolist.push(pcommand);
asswecan->undolist.top()->Execute();
}
cout << "Press (u) to undo and (r) to redo( (q) for quit): " << endl;
cin >> ch1;
}

delete fuck_u;
delete ddf;
delete pa;
delete asswecan;
return 0;
}


N-Queens-Problem

Without recursion, use backtracking to solve n queens problem.

#include <iostream>
#include <stdio.h>
#include <stack>
#define N 16
using namespace std;

class Operation {
int b[N][N];
int x;
int y;
stack <int> stack_alpha;
public:
Operation() {
for (int i = 0; i < N; i++) for (int j = 0; j < N; j++) b[i][j] = 0;
x = 0;
y = 0;
}

bool execute() {
for (; x < N; x++) {
if (is_Safe()) {
stack_alpha.push(x);
b[x][y] = 1;
y++;
x = 0;
return true;
}
}
return false;
}

bool is_Safe() {
int i ,j;
for (i = 0; i < y; i++) if (b[x][i]) return false;
for (i = x, j = y; i >= 0 && j >= 0; --i, --j) if (b[i][j]) return false;
for (i = x, j = y; j >= 0 && i < N; ++i, --j) if (b[i][j]) return false;
return true;
}

bool put_into_stack() {
do {
if (execute()) { }
else {
y--;
x = stack_alpha.top();
stack_alpha.pop();
cout << "x = " << x << endl;
b[x++][y] = 0;
}
if (y >= N) return true;
} while (x >= 0 && y >= 0);
return false;
}

void print_solution() {
if (put_into_stack()) {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
cout << b[i][j] << " ";
}
cout << endl;
}
} else cout << "not exist" << endl;
}

};

int main() {
Operation _operation;
_operation.print_solution();
}


using class Stack above.

class Operation {
int b[N+1][N+1];
int x;
int y;
int diag_a[2*MAXN];
int diag_b[2*MAXN];
short pre[MAXN];
short next[MAXN];
Stack stack_alpha;
int s;
int s1;
public:
Operation() {
for (int i = 1; i <= N; i++)
for (int j = 1; j <= N; j++) b[i][j] = 0;
memset(diag_a, 1, sizeof(diag_a));
memset(diag_b, 1, sizeof(diag_b));
for (int k = 1; k <= N; k++) {
pre[k] = k-1;
next[k] = k+1;
}
pre[N+1] = N;
next[0] = 1;
x = 0;
y = 1;
s = 0;
s1 = 0;
}

int put_into_stack() {
do {
if (stack_alpha.bot() <= N/2+1) {
if (y > N) {
if (stack_alpha.bot() == N/2+1) s1++;
else s++;
y--;
x = stack_alpha.peak_top();
diag_a[x+y] = 1;
diag_b[x-y+N] = 1;
next[pre[x]] = x;
pre[next[x]] = x;
b[x][y] = 0;
stack_alpha.pop();
x = stack_alpha.peak_top();
stack_alpha.pop();
y--;
}
else {
if (x != 0) {
diag_a[x+y] = 1;
diag_b[x-y+N] = 1;
next[pre[x]] = x;
pre[next[x]] = x;
}
x = next[x];
while (x <= N) {
if (diag_a[x+y] && diag_b[x-y+N]) break;
x = next[x];
}
if (x > N) {
y--;
x = stack_alpha.peak_top();
b[x][y] = 0;
stack_alpha.pop();
continue;
}
stack_alpha.push(x);
diag_a[x+y] = 0;
diag_b[x-y+N] = 0;
next[pre[x]] = next[x];
pre[next[x]] = pre[x];
b[x][y] = 1;
y++;
x = 0;
}
}
} while (stack_alpha.bot() <= N/2+1);
if (n%2 == 0) s*=2;
else s = 2*s + s1;
return s;
}
};

int main() {
Operation _operation;
int c = _operation.put_into_stack();
cout << c << endl;
}


for N = 16 it has 14772512 solutions.

Expression Evaluator

Based on the expression evaluator above, add × and ÷ support.

Based on the expression evaluator above, add brackets support.

#include <iostream>
#include <stack>
#include <cstring>
using namespace std;
#define N 100
class Caculator {
stack<double> num;
stack<char> op;
stack<double> pre;
char *s;
int k;
char c;
int end;
public:
Caculator(char _s[]) {
op.push('\0');
s = new char[strlen(_s) + 1];
strcpy(s, _s);
k = 0;
end = 0;
c = s[k];
}

double transform_num() {
int flag = 0;
double x = 0.0;
double y = 0.1;
while (s[k] >= '0' && s[k] <= '9' || s[k] == '.') {
if (s[k] >= '0' && s[k] <= '9') {
if (flag == 0) x = x*10 + (s[k] - '0');
else {
x = x + y*(s[k] - '0');
y*=0.1;
}
}
else flag = 1;
k+=1;
}
return x;
}
int priority(char opx) {
int k;
switch (opx) {
case '*' : k = 2; break;
case '/' : k = 2; break;
case '+' : k = 1; break;
case '-' : k = 1; break;
case '(' : k = 0; break;
case ')' : k = 0; break;
default : k = -1; break;
}
return k;
}

void caculate() {
int x, y;
cout << s << endl;
int i = 0;
while (!end) {
if (c >= '0' && c <= '9' || c == '.') {
num.push(transform_num());
}
else if (c == '(' || priority(c) > priority(op.top())) {
op.push(c);
k++;
}
else if (c == ')' && op.top() == '(') {
op.pop();
k++;
}
else if (c == 'a') {
num.push(pre.top());
k+=3;
}
else if (c == '\0' && op.top() == '\0') end = 1;
else if (priority(c) <= priority(op.top())) {
x = num.top();
num.pop();
y = num.top();
num.pop();
c = op.top();
op.pop();
switch (c) {
case '+' : y = x + y; break;
case '-' : y = y - x; break;
case '*' : y = x * y; break;
case '/' : y = x / y; break;
}
num.push(y);
}
c = s[k];
i++;
}
cout << "ans = " << num.top() << endl;
pre.push(num.top());
}
};

int main() {
char _s[100];
cin >> _s;
Caculator caculator(_s);
caculator.caculate();
int exit = 0;
};


Queue

implement a queue

Use arrays to implement queue.

• Enqueue a node into a given queue.

• Dequeue a node from a given queue.

• Empty a queue.

#include <stdio.h>
#include <iostream>
#include <limits.h>

struct Queue {
int front;
int rear;
int size;
unsigned capacity;
int* a;
};

Queue * create_queue(unsigned capacity) {
Queue * queue = (Queue*) malloc(sizeof(Queue));
queue->capacity = capacity;
queue->front = queue->size = 0;
queue->rear = capacity - 1;
queue->a = (int*) malloc(queue->capacity * sizeof(int));
return queue;
}

void enqueue(Queue * queue, int data) {
if (queue->size == queue->capacity) {
cout << "Queue overflow" << endl;
return;
}
queue->rear = (queue->rear+1)%queue->capacity;
queue->a[queue->rear] = data;
queue->size = queue->size + 1;
cout << "enqueued to queue " << data << endl;
}

int dequeue(Queue * queue) {
if (queue->size == 0) {
cout << "Queue underflow" << endl;
return;
}
int data = queue->a[queue->front];
queue->size = queue->size - 1;
return data;
}

int front(Queue * queue) {
return queue->a[queue->front];
}

int rear(Queue * queue) {
return queue->a[queue->rear];
}


Use pointers or references to implement linked lists.

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <iostream>
using namespace std;

struct Queue {
int data;
struct Queue* next;
};

void * push(Queue** bot, int value) {
StackNode * new_node = (StackNode*) malloc(sizeof(StackNode));
new_node->data = value;
new_node->next = *bot;
*bot = new_node;
cout << new_node->data << " pushed to stack" << endl;
}

int pop(StackNode** bot) {
if(!*bot) return -1;
StackNode* temp = *bot;
*bot = (*bot)->next;
int popped = temp->data;
free(temp);
return popped;
}

int peak(StackNode* top) {
if (!top) return -1;
for (element = bot; element != NULL && element->next != NULL; element = element->next) {}
return element;
}


Implement a circular buffer. // TODO: Better problem needed

Implement a message queue. // TODO: Better problem needed

Tree

Explain binary tree, full binary tree, complete binary tree.

In computer science, a binary tree is a tree data structure in which each node has at most two children, which are referred to as the left child and the right child.

A full binary tree (sometimes referred to as a proper[15] or plane binary tree)[16][17] is a tree in which every node has either 0 or 2 children.

In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2^h nodes at the last level h.

traversal of a tree

Given the root node of a tree, print its pre-order traversal, in-order traversal, post-order traversal and level-order traversal.

#include <iostream>
#include <stdlib.h>
using namespace std;

struct BiNode {
int data;
struct BiNode *l_child, *r_child;
};

class BiTree {
public:
void Create_Root(BiNode * root, int root_value) {
root->data = root_value;
}
void Create_BiTree(BiNode ** T);
bool is_Empty(BiNode* T);
int BiTree_Depth(BiNode* T);
void pre_Order(BiNode* T);
void in_Order(BiNode* T);
void post_Order(BiNode* T);
};

void BiTree :: Create_BiTree(BiNode ** T) {
int value;
cin >> (value);
if (!cin.good()) {
*T = NULL;
cin.clear();
cin.ignore();
}
else {
* T = (BiNode*) malloc(sizeof(BiNode));
(*T)->data = value;
Create_BiTree(&(*T)->l_child);
Create_BiTree(&(*T)->r_child);
}
}

bool BiTree :: is_Empty(BiNode* T) {
if(T) return false;
else return true;
}

int BiTree :: BiTree_Depth(BiNode* T){
int i, j;
if (T == NULL) return 0;
if (T->l_child) i = BiTree_Depth(T->l_child);
else i = 0;
if (T->r_child) j = BiTree_Depth(T->r_child);
else j = 0;
return i>j? i+1:j+1;
}

void BiTree :: pre_Order(BiNode* T) {
if (T != NULL) {
cout << T->data << " ";
pre_Order(T->l_child);
pre_Order(T->r_child);
}
}

void BiTree :: in_Order(BiNode* T) {
if (T != NULL) {
in_Order(T->l_child);
cout << T->data << " ";
in_Order(T->r_child);
}
}

void BiTree :: post_Order(BiNode* T) {
if (T != NULL) {
post_Order(T->l_child);
post_Order(T->r_child);
cout << T->data << " ";
}
}

int main() {
BiNode * root;
BiTree bitree;
//bitree.Create_Root(root, 1);
bitree.Create_BiTree(&root);
bitree.pre_Order(root);
}

• Same problem, without recursion.
void BiTree :: nr_pre_Order(BiNode* T) {
stack<BiNode*>lr;
BiNode * element = T;
cout << element->data << " ";
do {
if (element->l_child != NULL) {
lr.push(element);
cout << element->l_child->data << " ";
element = element->l_child;
}
else {
if (element->r_child != NULL) cout << element->r_child->data << " ";
do {
if (lr.empty()) return;
element = lr.top()->r_child;
lr.pop();
} while(element == NULL && !lr.empty());
if(element != NULL) {
cout << element->data << " ";
}
else return;
}
} while (true);
}


Given the post-order traversal and in-order traversal of a tree, print its pre-order traversal.

int BiTree :: find(int arr[], int start, int end, int value) {
int i;
for(i = start; i <= end; i++)
if (arr[i] == value) return i;
}

BiNode* BiTree :: reconstruction_i_p(int in[], int pre[], int in_start, int in_end) {
static int pre_i = 0;
if (in_start > in_end) return NULL;
BiNode* t_node = Create_BiNode(pre[pre_i++]);
if (in_start == in_end) return t_node;
int in_i = find(in, in_start, in_end, t_node->data);
t_node->l_child = reconstruction_i_p(in, pre, in_start, in_i-1);
t_node->r_child = reconstruction_i_p(in, pre, in_i+1, in_end);
return t_node;
}

BiNode* BiTree :: reconstruction_i_post(int in[], int post[], int in_start, int in_end) {
static int post_i = in_end;
if (in_start > in_end) return NULL;
BiNode* t_node = Create_BiNode(post[post_i--]);
if (in_start == in_end) return t_node;
int in_i = find(in, in_start, in_end, t_node->data);
t_node->r_child = reconstruction_i_post(in, post, in_i+1, in_end);
t_node->l_child = reconstruction_i_post(in, post, in_start, in_i-1);
return t_node;
}


Given a tree with a in-order traversal of which the data are in increasing order, i.e. BST, insert a new node while keeping this property.

• Implement a sorting algorithm with it (tree sort).
void BiTree :: insert_BiNode(BiNode* T, BiNode** last, int target) {
if (T != NULL) {
insert_BiNode(T->l_child, last, target);
if (flag == 1) return;
cout << T->data << endl;
if (target <= T->data) {
cout << target << " <= " << T->data << endl;
if ((*last)->r_child == NULL) {
BiNode* new_node = Create_BiNode(target);
(*last)->r_child = new_node;
} else {
BiNode* new_node = Create_BiNode(target);
T->l_child = new_node;
}
flag = 1;
return;
}
else cout << target << " > " << T->data << endl;
*last = T;
insert_BiNode(T->r_child, last, target);
}
}

BiNode* BiTree :: insert_BiNode(BiNode* T, int target) {
if (T == NULL) {
cout << "OK, root" << endl;
T = Create_BiNode(target);
return T;
}
if (target < T->data) T->l_child = insert_BiNode(T->l_child, target);
else T->r_child = insert_BiNode(T->r_child, target);
return T;
}

BiNode* BiTree :: insertion_sort(int d[], int n) {
BiNode* tree = NULL;
cout << "OK, start" << endl;
for (int i = 0; i < n; i++) tree = insert_BiNode(tree, d[i]);
return tree;
}


BST

Given n, how many structurally unique BST’s (binary search trees) that store values 1…n?

we can use Catalan Numbers to solve this problem.

The nth Catalan number is given directly in terms of binomial coefficients byC_{n} = \frac{n+1}{1}\bigl( \begin{smallmatrix} 2n\\n \end{smallmatrix} \bigr) = \frac{(2n)!}{(n+1)!n!} = \prod_{k=2}^{n}\frac{n+k}{k}

Analyze the complexity of BST, tell in which situation it behaves bad.

best case, the height h equals to O(n), which means complexity = O(logn)
because the time complexity depends on height of BST.

the worst case it behaves bad when BST degenerate to Single tree branch.

complexity = O(n)

Implement a binary heap.

• Consider a complete binary tree. Can it be properly stored in an array? How to get parent / child node of a given node?

absolutly.

one Node with index i, index of its left child2i+1 and right child2i+2

parent node (if exist) is
\frac{2}{i-1}

• If every node of this tree either has no parent (it is the root! ) or the datum of its parent is larger than its, it is called a heap. Can you insert a new node, and keep its properties (complete binary tree, parent datum larger than children datum)?
#include <iostream>
using namespace std;
#define N 1000;
class BiHeap {
int end;
int bh[N];
public:
BiHeap() {
end = 0;
cin >> bh[end];
}
int maxheap_insert() {
int data;
cin << data;
if (end == N) return 0;
bh[end] = data;
int c = end;
int p = (c-1)/2;
int temp = bh[c];
while(c > 0) {
if (bh[p] >= temp) break;
else {
bh[c] = bh[p];
c = p;
p = (p-1)/2;
}
}
bh[c] = temp;
end++;
return 1;
}
}

• if the root node is removed, can you transform the rest nodes into a heap?
    void delete_root() {
bh[0] = bh[size-1];
size--;
int c = 0;
int l = 2*c + 1;
int tmp = bh[c];
while (l < size) {
if (l < size-1 && bh[l] < bh[l+1]) l++;
if (tmp >= bh[l]) break;
else {
bh[c] = bh[l];
c = l;
l = 2*l + 1;
}
}
bh[c] = tmp;
}

• implement a sorting algorithm with it (heap sort).
void max_heapify(int a[], int start, int end) {
int son = 2*dad + 1;
while (son <= end) {
if (son <= end-1 && a[son] < a[son+1]) son++;
else {
}
}
}


diameter of a tree

Given a tree(no root node specified), print its diameter.

int BiTree :: diameter(BiNode* T) {
if (T == NULL) return 0;
int l_depth = BiTree_Depth(T->l_child);
int r_depth = BiTree_Depth(T->r_child);
int l_diameter = diameter(T->l_child);
int r_diameter = diameter(T->r_child);
return max(1+l_depth+r_depth, max(l_diameter, r_diameter));
}


Aho-Corasick Algorithm

Given a set of string, find them in a text.
Aho-Corasick Algorithm : complexity: O(n+m+z) where m is the total length of words and
z is the total number occurrences of words in text.

Huffman Tree

Construct Huffman tree with a given set of nodes and their weights.

#include <bits/stdc++.h>
#include <fstream>
using namespace std;
#define N 52

struct Node {
char data; //Characters
int weight;
struct Node *l_child, *r_child, *parent;
};

class HuffmanTree {
private:
vector<Node*>h_tree;
void count_num(); //count frequency of each character and the total number num
public:
HuffmanTree();
//void count_num(string &str);
};

void HuffmanTree :: count_num() {
int size = 52;
char ch[size];
int weight_tmp[size];
ch[0] = 'A';
ch[26] = 'a';
for (int i = 1; i < size/2; i++) {
ch[i] = ch[i-1] + 1;
ch[i+26] = ch[i+25] +1;
}
for (int i = 0; i < size; i++) weight_tmp[i] = 0;
for (int i = 0; i < s.size(); i++) {
if (s[i] > 64 && s[i] < 91) weight_tmp[s[i]-'A']++;
if (s[i] > 96 && s[i] < 123) weight_tmp[26+s[i]-'a']++;
}
for (int i = 0; i < size; i++) {
if(weight_tmp[i] != 0) {
Node* T = (Node*) malloc(sizeof(Node));
T->data = ch[i];
T->weight = weight_tmp[i];
T->parent = NULL;
T->l_child = NULL;
T->r_child = NULL;
h_tree.push_back(T);
cout << "done" << endl;
}
}
}

string str;
ifstream f1("test.txt",ios::in);
f1 >> str;
f1.close();
cout << str << endl;
cout << "OK2" << endl;
return str;
}
// customized compare-function of struct for sorting
bool comp(const Node* a, const Node* b) {
return a->weight > b->weight;
}
deque <Node*> c;  //定义一个空的队列
c.push_back(root);
while (!c.empty()) {  //如果队列不为空
Node* temp = c.front();  //返回队列的第一个元素
if (temp) {  //如果是非空结点
cout << temp->data << "-" << temp->weight << " ";
c.pop_front();  //出队列

c.push_back(temp->l_child);  //左孩子
c.push_back(temp->r_child); //右孩子
}
else {
c.pop_front();  //出队列
}
}
}

HuffmanTree :: HuffmanTree () {
count_num();
pair<Node*,Node*> small;
small.first = NULL;
small.second = NULL;
for (int i = 0; i < h_tree.size(); i++) {
cout << h_tree[i]->data << " " << h_tree[i]->weight << endl;
}
cout << "after sorting" << endl;
sort(h_tree.begin(),h_tree.end(),comp);
for (int i = 0; i < h_tree.size(); i++) cout << h_tree[i]->data << " " << h_tree[i]->weig
while(h_tree.size() != 1) {
Node* tmp_1 = h_tree.back();
h_tree.pop_back();
Node* tmp_2 = h_tree.back();
h_tree.pop_back();
Node* T = (Node*) malloc(sizeof(Node));
T->l_child = tmp_1;
T->r_child = tmp_2;
T->parent = NULL;
T->data = '0';
T->weight = tmp_1->weight + tmp_2->weight;
h_tree.push_back(T);
sort(h_tree.begin(),h_tree.end(),comp);
}
}
int main(){
new HuffmanTree();
return 0;
}



Graph

Store of Graph

using namespace std;
#define N 100 //max number of V
//add an edge in an undirected graph
}
void print_graph(vector<int> adj[], int n) {
for (int v = 0; v < n; v++) {
cout << "\n Adjacency list of vertex " << v << "\n head ";
for (auto x : adj[v]) cout << "-> " << x;
printf("\n");
}
}

int main() {
int n;
f1 >> n;
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++) f1 >> adj_m[i][j];
for (int i = 0; i < n; i++) {
for (int j  = 0; j < n; j++) {
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) cout << new_adj_m[i][j] << " ";
cout << endl;
}
}


result:

Adjacency list of vertex 0
head -> 1-> 4

Adjacency list of vertex 1
head -> 0-> 2-> 3-> 4

Adjacency list of vertex 2
head -> 1-> 3

Adjacency list of vertex 3
head -> 1-> 2-> 4

Adjacency list of vertex 4
head -> 0-> 1-> 3

0 1 0 0 1
1 0 1 1 1
0 1 0 1 0
0 1 1 0 1
1 1 0 1 0

• Explain the possible meaning of powers of adjacency matrices

If A is the adjacency matrix of the directed or undirected graph G, then the matrix A^{n} (i.e., the matrix product of n copies of A) has an interesting interpretation: the element (i, j) gives the number of (directed or undirected) walks of length n from vertex i to vertex j. If n is the smallest nonnegative integer, such that for some i, j, the element (i, j) of An is positive, then n is the distance between vertex i and vertex j. This implies, for example, that the number of triangles in an undirected graph G is exactly the trace of A^{3} divided by 6. Note that the adjacency matrix can be used to determine whether or not the graph is connected.

BFS/DFS Algorithm

Breadth-first search (BFS) is an algorithm for traversing or searching tree or graph data structures. It starts at the tree root (or some arbitrary node of a graph, sometimes referred to as a search key), and explores all of the neighbor nodes at the present depth prior to moving on to the nodes at the next depth level.
It uses the opposite strategy as depth-first search, which instead explores the highest-depth nodes first before being forced to backtrack and expand shallower nodes.WiKi

• pseudocode
BFS(G,s)
for\ each\ vertex\ u \in G.V- \lbrace s \rbrace
s.color = GRAY
s.d = 0
s. \pi = NIL
Q = \varnothing
ENQUEQUE(Q,s)
while\ Q \neq \varnothing
u = DEQUEUE(Q)
for\ each\ v \in G.adj[ u ]
$\qquad if\ v.color WHITE$
u.color = BLACK
total running time = O(V+E)

The breadth-first-search procedure BFS below assumes that the input graph
G = (V,E) is represented using adjacency lists. It attaches several additional
attributes to each vertex in the graph. We store the color of each vertex\ u \in V
in the attribute u.color and the predecessor of u in the attribute u. \pi. If u has no
predecessor (for example, if u D s or u has not been discovered), then\ u. \pi = NIL .
The attribute u.d holds the distance from the source s to vertex u computed by the
algorithm. The algorithm also uses a first-in, first-out queue Q (see Section 10.1)
to manage the set of gray vertices.(Introduction to Algorithms /Thomas Cormen /Leiserson)

implement with c++:

#include<bits/stdc++.h>
using namespace std;
struct Node {
int color; //0 represent white, 1-gray, 2-black
int d; //distence between node and root
Node* parent
};
vector<Node> V; // BFS-Tree

void BFS(vector<Node V>) {
n = V.size();
s = V[0];
for (int i = 1; i < n; i++) {
V[i].color = 0;
V[i].d = INT_MAX;
V[i].parent = NULL
}
s.color = 1;
s.d = 0;
s.parent = NULL;
queue<int> Q;
queue.push(0);
while(!Q.empty()) {
u = Q.front();
Q.pop();
if (V[x].color == 0) {
V[x].color = 1;
V[x].d = V[u].d + 1;
V[x].parent = &V[u];
Q.push(x);
}
}
V[u].color = 2;
}
}

• DFS

The strategy followed by depth-first search is, as its name implies, to search
“deeper” in the graph whenever possible. Depth-first search explores edges out
of the most recently discovered vertex v that still has unexplored edges leaving it.
Once all of v’s edges have been explored, the search “backtracks” to explore edges
leaving the vertex from which v was discovered. This process continues until we
have discovered all the vertices that are reachable from the original source vertex.
If any undiscovered vertices remain, then depth-first search selects one of them as
a new source, and it repeats the search from that source. The algorithm repeats this
entire process until it has discovered every vertex. (from "Introduction to Algorithms")

• pseudocode
DFG(G)
for\ each\ vertex\ u \in G.V
time = 0
for\ each\ vertex\ u \in G.V
$\qquad if u.color WHITE$
DFS-VISIT(G,u)
time = time + 1 //white vertex u has just been discovered
u.d = time
u.color = GRAY
for\ each\ v \in G:Adj[ u ] //explore edge (u,v)
u.color = BLACK
time = time + 1
u.f = time

process of DFS: initially set we all nodes to white, ....

#include<bits/stdc++.h>
using namespace std;

class Graph{
private:
int V;
void DFS_rec(int v, bool visited[]);
public:
Graph(int V);
void DFS(int v);
};

Graph :: Graph(int V) {
this->V = V;
}

void Graph :: add_edge(int v, int w){
}

void Graph :: DFS_rec(int v, bool visited[]) {
visited[v] = true;
cout << v << " ";
list<int>::iterator i;
if (!visited[*i]) DFS_rec(*i, visited);
}
}

void Graph :: DFS(int v) {
bool *visited = new bool[V];
for (int i = 0; i < V; i++) visited[i] = false;
DFS_rec(v, visited);
}

int main() {
int V;
f1 >> V;
for (int i = 0; i < V; i++)
for (int j = 0; j < V; j++) f1 >> adj_m[i][j];
Graph g(V);
for (int i = 0; i < V; i++)
for (int j = 0; j < V; j++)
g.DFS(0);
return 0;
}



SSC

• pseudocode
Strongly-Connected-Components(G)
1. $call\ DFS(G)\ to\ compute\ finishing\ times\ u.f\ for\ each\ vertex\ u$
2. $compute\ G^{T}$
3. $call\ DFS(G^{T},but\ in\ the\ main\ loop\ of\ DFS\ must\ consider \ the\ vertices\ in\ the\ decreased\ order\ of\ u.f)$
4. $output\ the\ vertices\ of\ each\ tree\ in\ the\ DF-forest \ which\ formed\ in\ line\ 3\ as\ a\ separate\ SSC$
void Graph :: get_trans() {
Graph g(V);
for (int v = 0; v < V; v++) {
list<int> :: iterator i;
}
}

void Graph :: creat_order(int v, bool visited[], stack<int> &Stack) {
visited[v] = true;
list<int> :: iterator i;
if (!visited[*i]) creat_order(*i, visited, Stack);
}
Stack.push_back(v);
}
void Graph :: print_SSC() {
stack<int> Stack;
bool *visited = new bool[V];
for (int i = 0; i< V; i++) visited[i] = false;
for (int i = 0; i < V; i++) {
if(visited[i] == false) creat_order(i, visited, Stack);
}
Graph gt = get_trans();
for (int i = 0; i < V; i++) visited[i] = false;
while(!Stack.empty()) {
int v = Stack.top();
Stack.pop();
if (visited[v] == false) {
gt.DFS_rec(v,visited);
cout << endl;
}
}
}


complexity: O(V+E)

• Generate minimum spanning tree of given graph.

Prim/Kruskal-Algorithm

Prim Algorithm

#include<bits/stdc++.h>
using namespace std;
#define N 5

int minKey(int key[],bool mst_set[]) {
int mini = INT_MAX;
int min_index;
for (int i = 0; i < N; i++) {
if (mst_set[i] == false && key[i] < mini) {
mini = key[i];
min_index = i;
}
}
return min_index;
}

void printMST(int parent[], int n, int adj_m[N][N]) {
printf("Edge \tWeight\n");
for (int i = 1; i < n; i++) printf("%d - %d \t%d \n", parent[i], i, adj_m[i][parent[i]]);
}
int parent[N];
int key[N];
bool mst_set[N];
for (int i  = 0; i < N; i++) {
key[i] = INT_MAX;
mst_set[i] = false;
}
key[0] = 0;
parent[0] = -1;
for (int cnt = 0; cnt < N-1; cnt++) {
int u = minKey(key, mst_set);
mst_set[u] = true;
for (int v = 0; v < N; v++) {
parent[v] = u;
}
}
}
}

int main() {
int n;
f1 >> n;
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++) f1 >> adj_m[i][j];
}


complexity:O(ElgV)

Kruskal-Algorithm
+ create a forest F (a set of trees), where each vertex in the graph is a separate tree
+ create a set S containing all the edges in the graph
+ while S is nonempty and F is not yet spanning
+ remove an edge with minimum weight from S
+ if the removed edge connects two different trees then add it to the forest F, combining two trees into a single tree
At the termination of the algorithm, the forest forms a minimum spanning forest of the graph. If the graph is connected, the forest has a single component and forms a minimum spanning tree

Pseudocode
1. MST-KRUSKAL(G,w)
2. Disjoin_set = \varnothing
3. for\ each\ vertex\ v \in G.V
5. sort\ the\ edge\ of\ G.E\ into\ nondecreasing\ order\ weight\ \omega
6. for\ each\ edge(u,v) \in G.E
7. \qquad if\ FIND-SET(v) \neq FIND-SET(u)
10. return A

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int> iPair;

bool comp(const pair<int,iPair> a, const pair<int,iPair> b) {
return a.first <= b.first;
}
struct Graph{
int V;
vector<pair<int,iPair>> edge;
int *parent, *rnk;
Graph(int V) {
this->V = V;
parent = new int[V];
rnk = new int[V];
for (int i  = 0; i <= V; i++) {
parent[i] = i;
rnk[i] = 0;
}
}
void add_edge(int w, int u, int v){
edge.push_back({w,{u,v}});
}
int MST_kruskal();
int find_p(int u){
if(parent[u] != u) parent[u] = find_p(parent[u]);
return parent[u];
}
void merge(int a,int b) {
a = find_p(a);
b = find_p(b);
if (rnk[a] > rnk[b]) parent[b] = a;
else parent[a] = b;
if (rnk[a] == rnk[b])  rnk[b]++;
}
};

int Graph :: MST_kruskal(){
int mst_w = 0;
sort(edge.begin(),edge.end(),comp);
vector<pair<int,iPair>> :: iterator it;
for (it = edge.begin(); it != edge.end(); it++) {
int u = it->second.first;
int v = it->second.second;
int set_u = find_p(u);
int set_v = find_p(v);
if (set_u != set_v) {
cout << u << " - " << v << endl;
mst_w += it->first;
merge(set_u,set_v);
}
}
return mst_w;
}

int main() {
int V;
f1 >> V;
Graph g(V);
for (int i = 0; i < V; i++) {
for (int j = 0; j < V; j++) {
int tmp;
f1 >> tmp;
if (tmp != 0 && j >= i) {
cout << i << " " << j << endl;
}
}
}
int ans = g.MST_kruskal();
cout << "\nWeight is " << ans << endl;;
return 0;
}


complexity: O(ElgV)

Dag-Shortest-Paths

with the relax-operation of each edge of a directed acyclic graph, which based on the topological order of nodes, we can calculate the shortest path between the source node and each rest node with time complexity O(V+E).
According to the algorithm, at first, we make topological sorting so that the linear order in nodes is fixed.

DAG-SHORTEST-PATHS(G,w,s)
topological sort the vetices of G
INITIALIZE-SINGLE-SOURCE(G,s)
for each vertex u, taken in topologically sorted order

struct node {
int vertex;
int d; //discovering time
int f; //time after finding
};

void comp(const node a, const node b) {
return a.f > b.f;
}
void Graph :: DSP() {
sort(topological_order,topological_order+V,comp);
for (int i = 0; i < V; i++) {
int u = tp[i].vertex;
list<int>::literator v;
if (way[*v] > way[u] + adj_m[u][*v]) {
parent[*v] = u;
}
}
}
}


time complexity: 'O(V+E)'

Dijkrsta-Algorithm

this algorithm requires weight of all edges be non-negative.
Dijkrsta-Algorithm keep a set of nodes as key information in process:the shortest path
from original node to each node of the set has been found. This algorithm repeatedly pick node
with shortest path and let it join to set S, then do RELAX on each node starting with u
in the following peusocode we use minimum-priority Queue to store the set.

DIJKRSTA(G,w,s)
INITIALIZE-SINGLE-SOURCE(G,s)
S = \varnothing
Q = G.V
while\ Q \neq \varnothing
unfinished->...