POJ 3259 Wormholes

While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ's farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

Input

Line 1: A single integer, F. F farm descriptions follow.
Line 1 of each farm: Three space-separated integers respectively: N, M, and W
Lines 2.. M+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.
Lines M+2.. M+ W+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.

Output

Lines 1.. F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).

Sample Input

2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8

Sample Output

NO
YES

Hint

For farm 1, FJ cannot travel back in time.
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this. 

Why POJ gg again ???

后来找了个别的OJ交的= =。。。

原题没有多组输入

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
struct node
{
    int u,v,w;
} eage[6000];
const int INF=0x3f3f3f3f;
int dis[6000];
int num;
int n,m,w;
void Add(int s,int e,int t)
{
    num++;
    eage[num].u=s;
    eage[num].v=e;
    eage[num].w=t;
}
int Bellman()//该算法用于图中有负权值的题目
{
    int i,j;
    int flag;
    for(i=0; i<=n; i++)
    {
        dis[i]=INF;
    }
    dis[1]=0;
    flag=0;
    for(i=1; i<=n; i++)
    {
        for(j=1; j<=num; j++)
        {
            if(dis[eage[j].v]>dis[eage[j].u]+eage[j].w)
            {
                dis[eage[j].v]=dis[eage[j].u]+eage[j].w;
            }
        }
    }
    for(i=1; i<=num; i++)
    {
        if(dis[eage[i].v]>dis[eage[i].u]+eage[i].w)
            return 1;
    }
    return 0;
}
int main()
{
    int t;
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            scanf("%d%d%d",&n,&m,&w);
            num=0;
            while(m--)
            {
                int s,e,t;
                scanf("%d%d%d",&s,&e,&t);
                Add(s,e,t);
                Add(e,s,t);
            }
            while(w--)
            {
                int s,e,t;
                scanf("%d%d%d",&s,&e,&t);
                Add(s,e,-t);
            }
            if(Bellman())printf("YES\n");
            else printf("NO\n");
        }
    }
    return 0;
}