Legal or Not （hdu 3342 拓扑排序判断是否成环）

Legal or Not

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

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
        

//题意：有师傅、徒弟关系，有些合法，有些非法，比如说1是2的师傅，2是3的师傅是合法的，但1是2的师傅，2是3的师傅，3是1的师傅就是非法的。题目还有句话是：还有如果1是2的师傅，2是3的师傅，那么1也是3的师傅。大概就这个意思，如果合法输出YES，不然输出NO。

//思路：就是很裸的判断图里是否存在环的问题。直接用个拓扑排序判断，拓扑排序如果把一个图里所有点都遍历到了就无环，反之有环。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;

const int MAX = 100 + 10;

int n, m;
vector<int>map[MAX];
int indegree[MAX];

bool toposort()
{
	int cnt = 0;
	queue<int>q;
	for (int i = 0; i < n; i++)
	{
		if (!indegree[i])
			q.push(i);
	}
	while (!q.empty())
	{
		int v = q.front();
		q.pop();
		for (int i = 0; i < map[v].size(); i++)
		{
			int u = map[v][i];
			indegree[u]--;
			if (!indegree[u])
				q.push(u);
		}
		cnt++;
	}
	if (cnt == n)
		return true;
	else
		return false;
}

int main()
{
	while (scanf("%d%d", &n, &m), n)
	{
		int x, y;
		memset(indegree, 0, sizeof(indegree));
		for (int i = 0; i <= n; i++)
			map[i].clear();
		for (int i = 0; i < m; i++)
		{
			scanf("%d%d", &x, &y);
			map[x].push_back(y);
			indegree[y]++;
		}
		if (toposort())
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}