POJ - 3259 I - 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 requiresT 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.

题意：一个农夫有n块农场，农场间有m条双向路，还有w个单向虫洞，可以回到之前的一个时间点，判断农夫可以在农场里碰到之前的自己吗。

分析：如果能遇到之前的自己，一定存在一个负环，用bellman_ford来判断是否有负环，把m个双向正边和w个单向负边存入，用bellman_ford松弛完后判断是否有负环。

代码；

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
int u[10010];
int v[10010];
int w[10010];
int dis[10010];
int n,m,k;
int solve()
{
    for(int i=1;i<=n;i++)
        dis[i]=INF;
    dis[1]=0;
    for(int i=1;i<=n;i++)
        for(int j=0;j<m*2+k;j++)
            if(dis[v[j]]>dis[u[j]]+w[j])  //松弛
                dis[v[j]]=dis[u[j]]+w[j];
    for(int i=0;i<m*2+k;i++)         //判断负环
        if(dis[v[i]]>dis[u[i]]+w[i])  //如果还能松弛  表示有负环 
            return 1;           
    return 0;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&k);
        for(int i=0;i<m*2;i++)  //存入双向正边
        {
            scanf("%d%d%d",&u[i],&v[i],&w[i]);
            i++;
            u[i]=v[i-1];
            v[i]=u[i-1];
            w[i]=w[i-1];
        }
        for(int i=m*2;i<m*2+k;i++)    //存入单向负边
        {
            scanf("%d%d%d",&u[i],&v[i],&w[i]);
            w[i]=-w[i];
        }
        if(solve())
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}