Harry and Magical Computer

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

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

       判断拓扑排序是否能够完成，拓扑排序不能完成的条件是没有环，只要判断图里是否有环是否就可以了，用flody和深搜应该都可以吧，我用的flody做的。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
int map[150][150];
int n,m;
void flody()
{
    for(int k=1;k<=n;k++)
       for(int i=1;i<=n;i++)
           for(int j=1;j<=n;j++)
               if(map[i][k]&&map[k][j])
                   map[i][j]=1;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(map,0,sizeof(map));
        int sign=0;
        int a,b;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&a,&b);
            if(a==b)
            {
                sign=1;
                continue;
            }
            map[a][b]=1;
        }
        if(sign)
        printf("NO\n");
        else
        {
            flody();
            for(int i=1;i<=n;i++)
            if(map[i][i])
            sign=1;
            if(sign)
            printf("NO\n");
            else
            printf("YES\n");
        }

    }
    return 0;
}