Hdoj 5154 Harry and Magical Computer 【拓扑】Harry and Magical Computer

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概述

Harry and Magical Computer

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1057 Accepted Submission(s): 430

Problem Description

In reward of being yearly outstanding magic student, Harry gets a magical computer. When the computer begins to deal with a process, it will work until the ending of the processes. One day the computer got n processes to deal with. We number the processes from 1 to n. However there are some dependencies between some processes. When there exists a dependencies (a, b), it means process b must be finished before process a. By knowing all the m dependencies, Harry wants to know if the computer can finish all the n processes.

Input

There are several test cases, you should process to the end of file.
For each test case, there are two numbers n m on the first line, indicates the number processes and the number of dependencies.

1≤n≤100,1≤m≤10000
The next following m lines, each line contains two numbers a b, indicates a dependencies (a, b).

1≤a,b≤n

Output

Output one line for each test case.
If the computer can finish all the process print "YES" (Without quotes).
Else print "NO" (Without quotes).

Sample Input

Sample Output


YES
NO

注意到：When there exists a dependencies (a, b), it means process b must be finished before process a. By knowing all the m dependencies, Harry wants to know if the computer can finish all the n processes.这一句，就是说b一定要在a前面完成，有一种先后关系，很明显拓扑。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
const int M = 105;
struct node{
int to, next;
}e[M*M];
bool vis[M];
int in[M], head[M*M];
int n, m, tot;
void init(){
/*for(int i = 0; i < M; ++ i){
/map[i].clear();
in[i] = 0;
vis[i] = 0;
}*/
memset(in, 0, sizeof(in));
memset(head, -1, sizeof(head));
}
void add(int a, int b){
e[tot].to = a;
e[tot].next = head[b];
head[b] = tot++;
}
bool topo(){
memset(vis, 0, sizeof(vis));
int cou = 0, num;
queue<int >q;
for(int i = 1; i <= n; ++i)
if(!in[i]){
vis[i] = 1;q.push(i);
}
while(!q.empty()){
int temp = q.front(); q.pop(); ++cou;
for(int i = head[temp]; ~i; i = e[i].next){
int v = e[i].to;
if(!(--in[v])&&!vis[v]){
q.push(v);
vis[v] = 1;
}
}
}
if(cou == n) return 1;
else return 0;
}
int main(){
while(scanf("%d%d", &n, &m) == 2){
init();
int a, b;
tot = 0;
for(int i = 0; i < m; ++ i){
scanf("%d%d", &a, &b);
add(a, b); ++in[a];
}
//cout << "sad"<<endl;
bool flag = topo();
if(flag) cout << "YESn";
else cout << "NOn";
}
return 0;
}