Problem Description

ACM-DIY is a large QQ group where many excellent acmers
get together. It is so harmonious that just like a big family. Every day,many
"holy cows" like HH, hh, AC, ZT, lcc, BF, Qinz and so on chat on-line to
exchange their ideas. When someone has questions, many warm-hearted cows like
Lost will come to help. Then the one being helped will call Lost "master", and
Lost will have a nice "prentice". By and by, there are many pairs of "master and
prentice". But then problem occurs: there are too many masters and too many
prentices, how can we know whether it is legal or not?

We all know a
master can have many prentices and a prentice may have a lot of masters too,
it's legal. Nevertheless，some cows are not so honest, they hold illegal
relationship. Take HH and 3xian for instant, HH is 3xian's master and, at the
same time, 3xian is HH's master,which is quite illegal! To avoid this,please
help us to judge whether their relationship is legal or not.

Please note
that the "master and prentice" relation is transitive. It means that if A is B's
master ans B is C's master, then A is C's master.

 

Input

The input consists of several test cases. For each
case, the first line contains two integers, N (members to be tested) and M
(relationships to be tested)(2 <= N, M <= 100). Then M lines follow, each
contains a pair of (x, y) which means x is y's master and y is x's prentice. The
input is terminated by N = 0.
TO MAKE IT SIMPLE, we give every one a number
(0, 1, 2,..., N-1). We use their numbers instead of their names.

 

Output

For each test case, print in one line the judgement of
the messy relationship.
If it is legal, output "YES", otherwise "NO".

 

Sample Input

 

Sample Output

 

Author

QiuQiu@NJFU

 

#include<stdio.h>

#include<string.h>

#include<queue>

using namespace std;

int n,m;

int map[110][110];

int vis[110];

void getmap()

{

    int i,j,a,b;

    memset(vis,0,sizeof(vis));

    memset(map,0,sizeof(map));

    for(i=1;i<=m;i++)

    {

        scanf("%d%d",&a,&b);

        if(!map[a][b])

        {

            vis[b]++;

            map[a][b]=1;

        }

    }

}

void tuopu()

{

    int i,j,sum=0;

    int ok=0;

    queue<int>q;

    while(!q.empty())

       q.pop();

    for(i=0;i<n;i++)

    {

        if(vis[i]==0)

        {

            sum++;

            q.push(i);

        }

    }

    if(sum==0) ok=1;//开始图中就不存在不依赖顶点的节点（成环）

    else

    {

        int u,ans=0;

        while(!q.empty())

        {

            u=q.front();

            ans++;

            q.pop();

            for(i=0;i<n;i++)

            {

                if(map[u][i])

                {

                    vis[i]--;

                    if(vis[i]==0)

                    q.push(i);

                }

            }

        }

        if(ans<n)//最后排序完成后不依赖顶点的节点个数小于n

            ok=1;//即存在环不符合题意

    }

    if(ok==0)

    printf("YES\n");

    else

    printf("NO\n");

}

int main()

{

    int i,j;

    while(scanf("%d%d",&n,&m),n|m)

    {

        getmap();

        tuopu();

    }

    return 0;

}