codeforces 915 D Almost Acyclic Graph

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

Input
3 4 1 2 2 3 3 2 3 1
Output
YES

Input
5 6 1 2 2 3 3 2 3 1 2 1 4 5
Output
NO

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.

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题意：给一个有向边的图，判断能否去掉一条边使其成为无环图。
思路：先dfs搜索一遍，把找到的第一个环返回；如果没有环打印YES。要成为无环图，那么这个环上一定要去掉一条边。所以只需要遍历这个环上所有的边，检查去掉后图上还有无其他的环，如果去掉一条边后图上无环，那么打印YES；如果去掉任何一条边都不能，打印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只检查有无环
    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，检查有无环，并返回遇到的第一个环
    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);//有环返回0
        pflag = rflag;
        g[ring[i]][ring[(i + 1) % nring]] = 1;

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