ural 1227 dfs判环&求最长路

问题可以简化为求最长路，因为边权均为正，所以有环即意味着最长路是正无穷必定有解；而无环的情况可以以每个点为起点进行dfs求出以该点为起点的最长路如果不小于s即有解。

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

const int INF = 999999999;
const int N = 101;
const int M = 20000;
bool visit[N];
int g[N][N];
int n, m, s;

int dfs( int u, int fa )
{
    int res = 0;
    visit[u] = 1;
    for ( int i = 1; i <= n; i++ )
    {
        if ( !g[u][i] ) continue;
        if ( i == fa ) continue;
        if ( visit[i] )
        {
            visit[u] = 0;
            return INF;
        }
        int tmp = dfs( i, u );
        if ( tmp == INF )
        {
            visit[u] = 0;
            return INF;
        }
        res = max( res, g[u][i] + tmp );
    }
    visit[u] = 0;
    return res;
}

int main ()
{
    while ( scanf("%d%d%d", &n, &m, &s) != EOF )
    {
        memset( g, 0, sizeof(g) );
        bool flag = false;
        while ( m-- )
        {
            int u, v, w;
            scanf("%d%d%d", &u, &v, &w);
            if ( g[u][v] || u == v )
            {
                flag = true;
            }
            g[u][v] = g[v][u] = w;
        }
        if ( flag )
        {
            puts("YES");
            continue;
        }
        memset( visit, 0, sizeof(visit) );
        for ( int i = 1; i <= n; i++ )
        {
            if ( dfs( i, -1 ) >= s )
            {
                flag = true;
                break;
            }
        }
        puts( flag ? "YES" : "NO" );
    }
    return 0;
}