POJ 3259 Wormholes

这题是一个模版题，可以借助bellman-ford算法中的判断负环的办法解决。

既然是模版题不多说上模版：

#include <stdio.h>

#define typec int // type of cost
const typec inf=0x3f3f3f3f; // max of cost
const int V=505;
const int E=8000;
int n, m, pre[V], edge[E][3];
typec dist[V];
int relax (int u, int v, typec c)
{
    if (dist[v] > dist[u] + c)
    {
        dist[v] = dist[u] + c;
        pre[v] = u;
        return 1;
    }
    return 0;
}
int bellman (int src)
{
    int i, j;
    for (i=0; i<n; ++i)
    {
        dist[i] = inf;
        pre[i] = -1;
    }
    dist[src] = 0;
    bool flag;
    for (i=1; i<n; ++i)
    {
        flag = false;
        for (j=0; j<m; ++j)
        {
            if( 1 == relax(edge[j][0], edge[j][1], edge[j][2]) ) flag = true;
        }
        if( !flag ) break;
    }
    for (j=0; j<m; ++j)
    {
        if (1 == relax(edge[j][0], edge[j][1], edge[j][2]))
            return 0;    //如果是负环返回0.
    }
    return 1;
}
int main()
{
    int F,M,W,u,v,w,i,j;
    scanf("%d",&F);
    while(F--)
    {
        scanf("%d%d%d",&n,&M,&W);
        m=0;
        for(i=0;i<M;i++)
        {
            scanf("%d%d%d",&edge[m][0],&edge[m][1],&edge[m][2]);
            m++;
            edge[m][0]=edge[m-1][1];
            edge[m][1]=edge[m-1][0];
            edge[m][2]=edge[m-1][2];
            m++;
        }
        for(i=0;i<W;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            w=-w;
            for(j=0;j<m;j++)
            {
                if(edge[i][0]==u&&edge[i][1]==v&&edge[i][2]>w)
                {
                    edge[i][2]=w;
                    break;
                }
            }
            if(j==m)
            {
                edge[m][0]=u;
                edge[m][1]=v;
                edge[m][2]=w;
                m++;
            }
        }
        if(bellman(1))
        {
            printf("NO\n");
        }
        else
        {
            printf("YES\n");
        }
    }
    return 0;
}