POJ 3259 - Wormholes

F - Wormholes

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  POJ 3259

Appoint description:  System Crawler  (2016-02-19)

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.

本质是判断是否存在负环，直接用bellman_ford，SPFA都行。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;

#define N 505

struct Edge{
    int u,v,t;
    Edge(int uu,int vv,int dd):u(uu),v(vv),t(dd){}
};

vector<Edge> edges;

int n,m,w;
int d[N];
#define INF 0x3f3f3f3f

bool bellman_ford(){
    for (int i=1;i<=n;i++)
        d[i]=INF;
    d[1]=0;

    for (int i=1;i<n;i++){
        bool flag=false;
        for (int j=0;j<edges.size();j++){
            int u,v,t;
            u=edges[j].u;
            v=edges[j].v;
            t=edges[j].t;
            if (d[u]<INF && d[u]+t<d[v]){
                d[v]=d[u]+t;
                flag=true;
            }
        }

        if (!flag)
            return false;
    }

    for (int j=0;j<edges.size();j++){
        int u,v,t;
        u=edges[j].u;
        v=edges[j].v;
        t=edges[j].t;
        if (d[u]<INF && d[u]+t<d[v]){
            d[v]=d[u]+t;
            return true;
        }
    }

    return false;
}

int main(){
    int T;
    scanf("%d",&T);
    while (T--){
        edges.clear();
        int u,v,d;
        scanf("%d%d%d",&n,&m,&w);
        for (int i=0;i<m;i++){
            scanf("%d%d%d",&u,&v,&d);
            edges.push_back(Edge(u,v,d));
            edges.push_back(Edge(v,u,d));
        }
        for (int i=0;i<w;i++){
            scanf("%d%d%d",&u,&v,&d);
            edges.push_back(Edge(u,v,-d));
        }

        if (bellman_ford()){
            printf("YES\n");
        }else{
            printf("NO\n");
        }
    }

    return 0;
}