poj 3259 Wormholes

                                                                                                      Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 45424		Accepted: 16771

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：农夫约翰想要通过虫洞回到过去，问有没有这么一种走法使得他能够回到过去。

思路：就是一个判断有无负圈的问题，存在负圈，说明能回到过去，可以选择用Bell_Ford算法。

AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
using namespace std;
const int N_MAX = 505,E_MAX=(2500+200+5)*2;
struct edge{
    int from, to, cost;
};
int d[N_MAX];
int V, E;
edge es[E_MAX];
bool find_negative_loop() {
    memset(d, 0, sizeof(d));
    for (int i = 0;i < V;i++) {
        for (int j = 0;j < E;j++) {
            edge e = es[j];
            if (d[e.to] > d[e.from] + e.cost) {
                d[e.to] = d[e.from] + e.cost;
                if (i == V - 1)return true;
            }
        }
    }
    return false;
}
int main() {
    int F,N,M,W;
    cin >> F;
    while (F--) {
        scanf("%d%d%d",&N,&M,&W);
        V = N;E =0;
        for (int i = 0;i < M;i++) {
            int from, to, cost;
            scanf("%d%d%d",&from ,&to,&cost);
            from--;
            to--;
            es[E].from = from;
            es[E].to = to;
            es[E].cost = cost;
            E++;
            es[E].from = to;//////无向图，双向的路
            es[E].to = from;
            es[E].cost = cost;
            E++;
        }
        for (int i = 0;i < W;i++) {
            int from, to, cost;
            scanf("%d%d%d", &from, &to, &cost);
            from--;
            to--;
            es[E].from = from;
            es[E].to = to;
            es[E].cost =-cost;
            E++;
        }
        if (find_negative_loop()) {
            printf("YES\n");
        }
        else printf("NO\n");
    }
    return 0;
}