You are given a directed graph consisting of n vertices and m edges (each edge is directed, so it can be traversed in only one direction). You are allowed to remove at most one edge from it.

Can you make this graph acyclic by removing at most one edge from it? A directed graph is called acyclic iff it doesn't contain any cycle (a non-empty path that starts and ends in the same vertex).

The first line contains two integers n and m (2 ≤ n ≤ 500,1 ≤ m ≤ min(n(n - 1), 100000)) — the number of vertices and the number of edges, respectively.

Then m lines follow. Each line contains two integers u and v denoting a directed edge going from vertex u to vertex v (1 ≤ u, v ≤ n,u ≠ v). Each ordered pair (u, v) is listed at most once (there is at most one directed edge fromu tov).

If it is possible to make this graph acyclic by removing at most one edge, print YES

3 4
1 2
2 3
3 2
3 1

YES

5 6
1 2
2 3
3 2
3 1
2 1
4 5

NO

In the first example you can remove edge , and the graph becomes acyclic.

In the second example you have to remove at least two edges (for example, and ) in order to make the graph acyclic.

        這題居然是topsort……

        大致題意就是給你一個圖，然後最短刪除一條邊，問你是否能夠把這個圖變成一個DAG有向無環圖。

        暴力的話列舉所有的邊，然後再遍歷圖，複雜度為O(M*(N+M))果斷超時。自己畫出圖來找一些環的個數以及相互之間的重邊的關係，還是沒有能夠找到一個很好的規律。另一個方法就是，找出一個的環，然後標記環上的邊，列舉這些邊刪掉，看是否還有環。但是這個操作起來不好處理，本身求具體的環就要用tarjan，然後還要在上面修改標記下環上的邊……

        正解是，既然刪掉一條邊對度的改變是是一個點入度減一，另一個點的出度減一，而拓撲排序正好可以判斷環而且與入度關聯，因此我們考慮列舉一個點，這個點關聯被刪掉的邊入度減一。我們把列舉的點入度減一，看看之後能否完整的做完拓撲排序，如果可以那麼就有解，否則沒有。那麼這個正確性如何理解呢？我們可以這樣想，少一條邊，相當於某個點入度減一，而這個點入度減一之後能否做完拓撲排序，則對應是否還有環，這樣就對應過來了。或者也可以理解為，列舉環上的點，讓他入度減一，相當於破掉與其關聯的一個環，再看是否有環。

        具體見程式碼：

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
#define LL long long
#define N 510

using namespace std;

int n,m,d[N],a[N];
vector<int> g[N];

bool topsort(int x)
{
    memcpy(a,d,sizeof(d));
    int num=0; a[x]--; queue<int> q;
    for(int i=1;i<=n;i++)
        if (a[i]==0) q.push(i),num++;
    while(!q.empty())
    {
        int i=q.front(); q.pop();
        for(int j=0;j<g[i].size();j++)
            if (!(--a[g[i][j]])) q.push(g[i][j]),num++;
    }
    return num==n;
}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        g[u].push_back(v); d[v]++;
    }
    for(int i=1;i<=n;i++)
        if (topsort(i))
        {
            puts("YES");
            return 0;
        }
    puts("NO");
    return 0;
}