在一個群裡面，大家互相請教問題，比如A請教B，我們就把B叫做師傅，把A叫做徒弟，這樣會產生很多“師傅——徒弟”的關系，一個徒弟可以有很多的師傅，一個師傅也可以有很多徒弟，這是合法的，但是不能出現A是B的師傅而且B是A的師傅，或者A是B的徒弟而且B是A的徒弟，或者在一個更大的關系環裡面出現這種情況。很明顯題目的意思就是，判斷一個給定的有向圖中是否存在環。了解了這些，解題方法就非常簡單了，那就是直接進行拓撲排序即可，統計拓撲排序完成之時能記錄的度為0的節點的個數，若個數等於節點個數則說明無環，否則是有環的。

Legal or Not

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4713 Accepted Submission(s): 2148


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

這個拓撲序的思路就是在訪問圖中一個節點後就將其後繼節點的入度減1，這樣若x節點的入度為0，則說明x節點可以在接下來被訪問。在這裡借助棧來保存入度為0且還未被訪問的節點。

#include 
#include 
#include 

using namespace std;

const int N = 105;

vector adj[N];
int in_cnt[N];
stack in_zero_st;

bool top_sort(int n)
{
	int cnt = n, u, v;
	while (!in_zero_st.empty())
	{
		u = in_zero_st.top();
		--cnt;
		in_zero_st.pop();

		for (int i = 0; i < adj[u].size(); ++i)
		{
			v = adj[u][i];
			--in_cnt[v];
			if (in_cnt[v] == 0)in_zero_st.push(v);
		}
	}

	if (cnt == 0)return true;
	return false;
}

int main()
{
	int n, m, u, v;

	while (scanf("%d %d", &n, &m))
	{
		if (n == 0)break;

		memset(in_cnt, 0, sizeof(in_cnt));

		for (int i = 0; i < m; ++i)
		{
			scanf("%d %d", &u, &v);
			adj[u].push_back(v);
			++in_cnt[v];
		}

		for (int i = 0; i < n; ++i)
		{
			if (in_cnt[i] == 0)in_zero_st.push(i);
		}

		//true indicates legal false for the other
		bool ret = top_sort(n);

		if (ret)
		{
			printf("YES\n");
		}
		else
		{
			printf("NO\n");
		}

		for (int i = 0; i < n; ++i)
		{
			adj[i].clear();
		}
	}

	return 0;
}