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题目1448：Legal or Not

时间限制：1 秒

内存限制：128 兆

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 20 11 22 20 11 00 0

样例输出：

YESNO

代码如下：

#include <stdio.h>#include <vector>#include <queue>using namespace std;#define N 101vector<int> edge[N];queue<int> q;int main(){int n, m;int inDegree[N];  // 统计每个结点的入度while (scanf("%d", &n) != EOF){if (n == 0) break;scanf("%d", &m);for (int i = 0; i < n; i++){inDegree[i] = 0;edge[i].clear();}while (!q.empty()) q.pop();while (m--){int a, b;scanf("%d%d", &a, &b);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);}for (int i = 0; i < n; i++)if (0 == inDegree[i]) q.push(i);  // 入度为0的结点入队int cnt = 0;while (!q.empty()){int nowP = q.front();q.pop();cnt++;for (int i = 0; i < edge[nowP].size(); i++)if (0 == --inDegree[edge[nowP][i]]) q.push(edge[nowP][i]);}puts(cnt == n ? "YES" : "NO");}return 0;}