拓扑排序例题——判断有环和无环

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

题意：本题的意思就是判断拓扑排序是否有环
本题我采用邻接矩阵作图。

#include<stdio.h>
#include<string.h>
const int M=1010;
int map[M][M],book[M],f[M],s[M];//map画图，book标记使用过的入度为零的点
int n,m,s1;						//f记录入度，s存储入度为零的点
void tuopu()
{
    int sum=0,i,j;
    while(sum<n)
    {
        s1=0;
        for(i=0;i<n;i++)
        {
            if(book[i]==0&&f[i]==0)//此点未用过且入度为零
            {
                s1=1;
                book[i]=1;
                break;
            }
        }
        if(s1==0)//表明没有入度为零的点，说明有环
        {
            printf("NO\n");
            return;
        }
        sum++;
        for(j=0;j<n;j++)//将以i为起点的边删除并且使终点的入度减一
        {
            if(map[i][j]==1)
            {
                map[i][j]=0;
                f[j]--;
            }
        }
    }
}
int main()
{
    while(~scanf("%d %d",&n,&m)&&n!=0||m!=0)
    {
        int i,j,k,a,b;
        memset(book,0,sizeof(book));
        memset(f,0,sizeof(f));
        memset(s,0,sizeof(s));
        memset(map,0,sizeof(map));
        for(i=1;i<=m;i++)
        {
            scanf("%d %d",&a,&b);
            if(map[a][b]==0)//目的，防止重边使得入度加一导致错误
            {
                map[a][b]=1;
                f[b]++;
            }
        }
        tuopu();
        if(s1==1)
            printf("YES\n");
    }
    return 0;
}