hdu 1599 无向图最小环

裸的无向图最小环，我们可以在Floyd的过程中求得。

具体思想是：枚举环中的编号最大的顶点k，然后枚举和k相关联的两个顶点i和j（i和j都小于k），则该环的长度为：dist[i][j] + g[j][k] + g[k][i]，循环到k的时候，dist[i][j]表示的是i到j的经过的顶点编号小于k的最短路，所以正确性很显然，由于i、j、k都不相同，所以环上至少是3个点。

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

const int INF = 999999;
const int N = 101;
int dist[N][N];
int g[N][N];
int n, m, ans;

void floyd()
{
    for ( int i = 1; i <= n; i++ )
    {
        for ( int j = 1; j <= n; j++ )
        {
            dist[i][j] = g[i][j];
        }
        dist[i][i] = g[i][i] = 0;
    }
    ans = INF;
    for ( int k = 1; k <= n; k++ )
    {
        for ( int i = 1; i < k; i++ )
        {
            for ( int j = i + 1; j < k; j++ )
            {
                int tmp = dist[i][j] + g[i][k] + g[k][j];
                if ( tmp < ans ) ans = tmp;
            }
        }
        for ( int i = 1; i <= n; i++ )
        {
            for ( int j = 1; j <= n; j++ )
            {
                int tmp = dist[i][k] + dist[k][j];
                if ( tmp < dist[i][j] )
                {
                    dist[i][j] = tmp;
                }
            }
        }
    }
}

int main ()
{
    while ( scanf("%d%d", &n, &m) != EOF )
    {
        for ( int i = 1; i <= n; i++ )
        {
            for ( int j = 1; j <= n; j++ )
            {
                g[i][j] = INF;
            }
        }
        while ( m-- )
        {
            int a, b, c;
            scanf("%d%d%d", &a, &b, &c);
            if ( c < g[a][b] )
            {
                g[a][b] = g[b][a] = c;
            }
        }
        floyd();
        if ( ans == INF )
        {
            printf("It's impossible.
");
        }
        else
        {
            printf("%d
", ans);
        }
    }
    return 0;
}