poj 3259 Wormholes

和poj 1860差不多都用bellman—ford最简单的版本就可以了，关键在于巧妙的运用，做过了才能更好地理解》

#include<stdio.h>
#include<queue>
#include<string.h>
#define INF 100000000
#define MAXN1 520
#define MAXN 8000

int u[MAXN],v[MAXN],wt[MAXN], d[MAXN1];
int n, m, w, T,p;

void init()
{
    scanf("%d%d%d",&n,&m,&w);
    p = 0;
    for(int i = 0; i < m; i ++)
    {
        scanf("%d%d%d",&u[p],&v[p],&wt[p]);
        p++;
        u[p] = v[p-1];
        v[p] = u[p-1];
        wt[p] = wt[p-1];
        p++;
    }
    for(int i = m; i < m+w; i ++)
    {
        scanf("%d%d%d",&u[p],&v[p],&wt[p]);
        wt[p] = -wt[p];
        p++;
    }
}

void bellman_ford(int h)
{
    for(int i = 1; i <= n; i ++) d[i] = 0;
    d[h] = 0;
    for(int k = 0; k < n-1; k ++)
    for(int i = 0; i < p; i ++)
    {
        int x = u[i], y = v[i];
        if(d[x] < INF && d[y] > d[x] + wt[i])
        d[y] = d[x] + wt[i];
    }
}

int main()
{
    while(~scanf("%d",&T))
    {
        while(T --)
        {
            init();
            int ok = 0;
                bellman_ford(1);
                for(int j = 0; j < p; j ++)
                {
                    int x = u[j], y = v[j];
                    if(d[x] < INF && d[y] > d[x] + wt[j])
                    {ok = 1; break;}
                }
            if(ok == 1) printf("YES\n");
            else printf("NO\n");
        }
    }
    return 0;
}