D. Almost Acyclic Graph(有向图找环+判环)

题目

题意：

    给定一张有向图，判断是否可以删除一条边使得这张图没有环。
     $2 ≤ n ≤ 500, 1 ≤ m ≤ min(n(n - 1), 100000)$

分析：

    思路很简单，随便找一个环，要使得无环，那么必然删除这个环上的某一条边。暴力删除判断是否还有环。对于有向图的找环，需要用两个数组。visx[i]表示以i为起点都搜索过了，vis[i]表示在本次的搜索树中i已经被遍历过了，每次返回上一层需要把vis[i]置为1。

#include <iostream>
#include <vector>
#include <cstring> 
using namespace std;

struct edge{
	int x,y;
	edge(int a,int b)
	{
		x = a;
		y = b;
	}
};

vector<int> g[505];
vector<edge> e;

int flag = 0,flagx = 0,vis[505],visx[505];

void dfs(int x)
{
	if( vis[x] == 1 )
	{
		flag = x;
		return;
	}
	if( visx[x] == 1 ) return;
	visx[x] = 1;
	vis[x] = 1;
	for (int i = 0; i < g[x].size(); i++)
	{
		int t = g[x][i];
		dfs(t);
		if( flagx ) 
		{
			vis[x] = 0;
			return;
		}
		if( flag )
		{
			e.push_back(edge(x,t)); 
			if( x == flag ) flagx = 1;
			vis[x] = 0;
			return; 
		}
	}
	vis[x] = 0;
}

void dfs2(int x,int index)
{
	visx[x] = 1;
	vis[x] = 1;
	for (int i = 0; i < g[x].size(); i++)
	{
		int t = g[x][i];
		if( e[index].x == x && e[index].y == g[x][i] ) continue;
		if( vis[t] )
		{
			flagx = 1;
			vis[x] = 0;
			return;
		}
		if( visx[t] ) continue;
		dfs2(g[x][i],index);
		if( flagx ) 
		{
			vis[x] = 0;
			return;
		}
	}
	vis[x] = 0;
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n,m;
	cin >> n >> m;
	for (int i = 1; i <= m; i++)
	{
		int x,y;
		cin >> x >> y;
		g[x].push_back(y); 
	}
	for (int i = 1; i <= n; i++)
	{
		flag = 0,flagx = 0;
		if( visx[i] == 0 ) dfs(i);
		if( flag ) break;
	}
	for (int i = 0; i < e.size(); i++)
	{
		flagx = 0;
		memset(visx,0,sizeof(visx));
		for (int j = 1; j <= n; j++)
		{
			if( visx[j] == 0 ) dfs2(j,i);
			if( flagx == 1 ) break;
		}
		if( flagx == 0 )
		{
			cout << "YES" << '\n';
			return 0;
		}
	}
	if( e.empty() ) cout << "YES" << '\n';
	else cout << "NO" << '\n';
	return 0;
}