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.

有n个点，m条双向边时间花费为c，w条单向边时间花费为-c，求农夫是否可以回到起点时，自己还没出发。即求是否存在负权环，用floyd直接求出了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define inf 0x3f3f3f3f

using namespace std;

int paths[510][510];
int n,m,k;

int floyd()
{
    for(int k=1;k<=n;k++)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(paths[i][j]>paths[i][k]+paths[k][j])
                {
                    paths[i][j]=paths[i][k]+paths[k][j];//更新路径长度
                }
            }
            if(paths[i][i]<0)return 1;
        }
    }
    return 0;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&k);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(i==j)paths[i][j]=0;
                else paths[i][j]=inf;
            }
        }
        int a,b,c;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(paths[a][b]>c)
            {
                paths[a][b]=paths[b][a]=c;
            }
        }
        for(int i=1;i<=k;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            paths[a][b]=-c;//在虫洞的时间为负
        }
        int ans=floyd();
        if(ans)
        {
            printf("YES\n");
        }
        else
        {
           printf("NO\n");
        }
    }
    return 0;
}