Wormholes POJ - 3259

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#最短路

本文介绍了一个有趣的编程问题：如何通过特定的路径和虫洞让农夫约翰实现时间旅行回到过去。该问题通过Bellman-Ford算法解决，判断是否存在负权回路使起点到终点的时间为负。

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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 backT 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

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.

Bellman-ford负权回路,一开始没把双虫洞看成双向

边，改完后数组忘了扩大，唉。

#include <iostream>
#include <cstdio>

using namespace std;

typedef long long ll;

const int M=6000+5;
const int INF=100000;

int from[M];
int to[M];
int cost[M];
int dis[M];

void solve(){
    int f;
    cin>>f;
    while(f--){
        int n,m,w;
        cin>>n>>m>>w;

        for(int i=1;i<=2*m;i+=2){
            int x,y,z;
            cin>>x>>y>>z;
            from[i]=x;
            to[i]=y;
            cost[i]=z;

            from[i+1]=y;
            to[i+1]=x;
            cost[i+1]=z;
        }
        for(int i=1;i<=w;i++){
            int x,y,z;
            cin>>x>>y>>z;
            from[i+2*m]=x;
            to[i+2*m]=y;
            cost[i+2*m]=-z;
        }

        for(int i=1;i<=n;i++){
            dis[i]=INF;
        }
        dis[1]=0;

        bool ff;
        for(int k=1;k<n;k++){
            ff=1;
            for(int i=1;i<=2*m+w;i++){
                if(dis[to[i]]>dis[from[i]]+cost[i]){
                    dis[to[i]]=dis[from[i]]+cost[i];
                    ff=0;
                }
            }
            if(ff)break;
        }

        bool flag=0;
        for(int i=1;i<=2*m+w;i++){
            if(dis[to[i]]>dis[from[i]]+cost[i]){
                flag=1;
            }
        }
        if(flag==1){
            cout<<"YES"<<endl;
        }
        else{
            cout<<"NO"<<endl;
        }

    }

}

int main(){
    solve();
    return 0;
}