题目1448：Legal or Not

题目描述：

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.

输入：

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.

输出：

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

样例输入：

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

样例输出：

YES
NO

大意：在一个qq群里有着许多师徒关系，如A是B的师父，同时B是A的徒弟，一个师父也可能会有许多不同的师父。输入给出该群里所有的师徒关系，问是否存在这样一种非法情况：以三人为例，即A是B的师父，B是C的师父，C又反过来是A的师父。若我们将该群里的所有人都抽象为图上的结点，将所有的师徒关系都抽象为有向边（由师父指向徒弟），该实际问题就转化为一个数学问题——该图上是否存在一个环，即判断该图是否为有向无环图

#include <stdio.h>
#include <vector>
#include <queue>
using namespace std;
vector<int>edge[501];//邻接链表，因为边不存在权值，只需保存与其邻接的结点编号即可，所以vector中的元素为int
queue<int>Q;//保存入度为0的队列

int main()
{
	int inDegree[501];//统计每个结点的入度
	int n, m;
	while (scanf("%d%d", &n, &m) != EOF) {
		if (n == 0 && m == 0)break;
		for (int i = 0;i < n;i++) {//初始化所有结点，注意本体结点编号由0到n-1
			inDegree[i] = 0;	   //初始化入度信息，所有结点入度均为0
			edge[i].clear();	   //清空邻接链表
		}
		while (m--) {
			int a, b;
			scanf("%d%d", &a, &b);//读入一条由a指向b的有向边
			inDegree[b]++;		  //又出现了一条弧头指向b的边，累加结点b的入度
			edge[a].push_back(b); //将b加入a的邻接链表
		}
		while (Q.empty() == false)Q.pop();//若队列非空，则一直弹出队头元素，该操作的目的为清空队列中的所有的元素（可能为上一组测试数据中遗留的数据）
		for (int i = 0;i < n;i++) {       //统计所有结点的入度
			if (inDegree[i] == 0)Q.push(i);//若结点入度为0，则将其放入队列
		}
		int cnt = 0;			  //计数器，初始值为0，用于累加已经确定拓扑序列的结点个数
		while (Q.empty() == false) {//当队列中入度为0的结点未被取完时重复
			int nowP = Q.front();//读出队头结点编号，本例不需要求出确定的拓扑序列，固不做处理；若要求求出确定的拓扑次序，则将该结点紧接着放在已经确定的拓扑序列之后
			Q.pop();//弹出队头元素
			cnt++;//被确定的结点个数加一
			for (int i = 0;i < edge[nowP].size();i++) {//将该结点以及以其为护卫的所有边去除
				inDegree[edge[nowP][i]]--;//去除某条边后，该边所指后继结点入度减一
				if (inDegree[edge[nowP][i]] == 0) {//若该结点入度变为0
					Q.push(edge[nowP][i]);//将其放入队列当中
				}
			}
		}
		if (cnt == n)puts("YES");//若所有结点都能被确定拓扑序列，则原图为有向无环图
		else puts("NO");//否则，原图为非有向循环图
	}
    return 0;
}