Description

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).

Input

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 from u to v).

Output

If it is possible to make this graph acyclic by removing at most one edge, print YES. Otherwise, print NO.

Examples

Note

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.

題意：給一個有向邊的圖，判斷能否去掉一條邊使其成為無環圖。
思路：先dfs搜尋一遍，把找到的第一個環返回；如果沒有環列印YES。要成為無環圖，那麼這個環上一定要去掉一條邊。所以只需要遍歷這個環上所有的邊，檢查去掉後圖上還有無其他的環，如果去掉一條邊後圖上無環，那麼列印YES；如果去掉任何一條邊都不能，列印NO。

#include<stdio.h>
//#include<iostream>
//using std::cin;
//using std::cout;
//#include<algorithm>
//using std::sort;
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define mabs(x) ((x)>0?(x):(0-(x)))
#define N_max 505
#include<memory.h>
int n,m;
int g[N_max][N_max] = { 0 };

int vis[N_max] = { 0 };
int ring[N_max] = { 0 };
int nring = 0;

int chk[N_max] = { 0 };
int checkring(int cur) {//dfs只檢查有無環
    if (chk[cur] == -1)
        return 0;

    chk[cur] = -1;
    for (int next = 1; next <= n; ++next) {
        if (chk[next] != 1 && g[cur][next] == 1) {
            if (0== checkring(next)) {
                return 0;
            }
        }
    }
    chk[cur] = 1;
    return 1;
}

int findring(int cur,int rdx) {//dfs，檢查有無環，並返回遇到的第一個環
    if (vis[cur] == -1) { 
        ring[rdx] = cur; nring = rdx;
        return 1; 
    }
    vis[cur] = -1; 
    for (int next = 1; next <= n; ++next) {
        if (vis[next]!=1&&g[cur][next]==1) {
            if (1 == findring(next,rdx+1)) { 
                ring[rdx] = cur;
                return 1; 
            }
        }
    }
    vis[cur] = 1;
    return -1;
}
int main() {
    scanf("%d %d", &n,&m);
    int a1, a2;
    for (int i = 0; i < m; ++i) {
        scanf("%d %d", &a1, &a2);
        g[a1][a2] = 1;
    }

    for (int i = 1; i <= n; ++i)
    {
        nring = 0;
        if (1 == findring(i,0)) {
            break;
        }
    }
    int pflag = 0,rflag;
    if (nring == 0) {
        printf("YES"); return 0;
    }
    ++nring;
    for (int i = 0; i<nring&&pflag==0; ++i) {
        memset(chk, 0, sizeof chk);
        g[ring[i]][ring[(i + 1) % nring]] = 0;
        rflag = 1;
        for (int t = 1; t <= n&&rflag;++t)
            rflag = checkring(t);//有環返回0
        pflag = rflag;
        g[ring[i]][ring[(i + 1) % nring]] = 1;

    }
    if (pflag)printf("YES");
    else printf("NO");
}