HDU5154拓扑排序

Harry and Magical Computer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2559    Accepted Submission(s): 978

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

3 2

3 1

2 1

3 3

3 2

2 1

1 3

 

Sample Output

YES

NO

 

Source

BestCoder Round #25

 题意i：

n个点，m条有向边，问没有环输出yes,有环输出no.

代码：

//直接，拓扑排序然后判断是否还有没入队的点就行。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
int n,m,in[],cnt[],nu;
vector<int>g[];
void topu()
{
    queue<int>q;
    nu=;
    for(int i=;i<=n;i++)
        if(in[i]==) q.push(i);
    while(!q.empty()){
        int u=q.front();q.pop();
        cnt[++nu]=u;
        for(int i=;i<g[u].size();i++){
            int p=g[u][i];
            if(--in[p]==)
                q.push(p);
        }
    }
}
int main()
{
    while(scanf("%d%d",&n,&m)==){
        for(int i=;i<=n;i++){
            in[i]=;
            g[i].clear();
        }
        int x,y;
        for(int i=;i<m;i++){
            scanf("%d%d",&x,&y);
            g[x].push_back(y);
            in[y]++;
        }
        topu();
        if(nu>=n) printf("YES\n");
        else printf("NO\n");
    }
    return ;
}