hdu 3342 拓扑排序

Legal or Not

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4796    Accepted Submission(s): 2198

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

3 2
0 1
1 2
2 2
0 1
1 0
0 0

 

Sample Output

YES
NO
题目大意：在一个群里面，大家互相请教问题，比如A请教B，我们就把B叫做master，把A叫做prentice，这样会产生很多master——prentice的关系，一个prentice可以有很多的master，一个master也可以有很多prentice，这是合法的，但是不能出现A是B的master而且B是A的master，或者在一个更大的关系环里面出现这种情况。

其实，问题就是判断一个有向图中是否有环的出现（拓扑排序）！！
#include<stdio.h>
#include<queue>
#include<cstring>
#define MAXN 105
using namespace std;
int cnt;
int map[MAXN][MAXN],into[MAXN],list[MAXN];
int main()
{
    queue <int> q;
    int dot,m,n,t,x,y;
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        cnt=0;
        while (!q.empty())
            q.pop();
        if (n==0&&m==0)
            break;
        memset(map,0,sizeof(map));
        memset(into,0,sizeof(into));
        for (int i=0; i<m; i++)
        {
            scanf("%d%d",&x,&y);
            map[x][++map[x][0]]=y;
            into[y]++;
        }
        for (int i=0; i<n; i++)
            if (into[i]==0)
                q.push(i);
        while (!q.empty())
        {
            dot=q.front();
            for (int i=1; i<=map[dot][0]; i++)
            {
                into[map[dot][i]]--;
                if (into[map[dot][i]]==0)
                    q.push(map[dot][i]);
            }
            list[cnt++]=dot;
            q.pop();
        }
        if (cnt==n)
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}