九度1448：Legal or Not

題目描述：

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.

輸入：

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.

輸出：

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

樣例輸入：

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

樣例輸出：

YES
NO

拓撲排序問題。代碼如下：

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
const int MAXN = 501;
vector<int> edge[501];
queue<int> que;
vector<int> coll;//確定的拓撲序列
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("D:\\in.txt", "r", stdin);
    freopen("D:\\out.txt", "w", stdout);
#endif
    int indegree[MAXN];//統計每個結點的入度
    int n(0), m(0);
    while (scanf("%d%d", &n, &m) != EOF)
    {
        if (n == 0 && m == 0)
            break;
        fill(indegree, indegree + MAXN, 0);
        for (int i = 0; i < n; i++)
        {
            edge[i].clear();//清空鄰接鏈表
        }
        while (m--)
        {
            int a(0), b(0);
            scanf("%d%d", &a, &b);//讀入一條由a指向b的有向邊
            indegree[b]++;//這邊邊指向b,b的入度加一
            edge[a].push_back(b);//將b加入a的鄰接鏈表
        }
        while (!que.empty())
            que.pop();
        for (int i = 0; i < n; i++)
        {
            if (indegree[i] == 0)
                que.push(i);
        }
        int cnt = 0;//已經確定拓撲序列的結點個數
        while (!que.empty())
        {
            int now = que.front();
            que.pop();
            cnt++;
            for (int i = 0; i < edge[now].size(); i++)
            {
                indegree[edge[now][i]]--;
                if (indegree[edge[now][i]] == 0)
                    que.push(edge[now][i]);
            }
        }
        if (cnt == n)
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}