poj 3259 Wormholes

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24061		Accepted: 8576

Description

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.

题意：一个农场里有若干虫洞，问是否能通过虫洞使自己回到出发之前的时间。

思路：判断图中有没有负权环。用Bellman-Ford算法。

AC代码：

#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <map>
using namespace std;
const int INF=10000000;
struct Edge
{
    int s,e;
    int t;
}edges[6000];
int n,m,w,cnt;
int dis[1005];

bool Bellman_ford()
{
    for(int i=1;i<=n;i++)
    dis[i]=INF;
    for(int i=1;i<=n-1;i++)
    {
        bool flag=false;
        for(int j=0;j<cnt;j++)
        {
            if(dis[edges[j].e]>dis[edges[j].s]+edges[j].t)
            {
                dis[edges[j].e]=dis[edges[j].s]+edges[j].t;
                flag=true;
            }
        }
        if(!flag) break;
    }
    for(int i=0;i<cnt;i++)
    if(dis[edges[i].e]>dis[edges[i].s]+edges[i].t)
    return true;
    return false;
}
int main()
{
    int t,l,u,v;
    cin>>t;
    while(t--)
    {
        cin>>n>>m>>w;
        cnt=0;
        while(m--)
        {
            cin>>u>>v>>l;
            edges[cnt].s=edges[cnt+1].e=u;
            edges[cnt].e=edges[cnt+1].s=v;
            edges[cnt++].t=l;
            edges[cnt++].t=l;
        }
        while(w--)
        {
            cin>>u>>v>>l;
            edges[cnt].s=u;
            edges[cnt].e=v;
            edges[cnt++].t=-l;
        }
        if(Bellman_ford())
        cout<<"YES"<<endl;
        else
        cout<<"NO"<<endl;
    }
    return 0;
}