POJ3259 Wormholes

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.

 

『题解』

存在负权边时不能用Dijkstra算法求最短路。本题采用Bellman-Ford最短路算法判断负权回路。模板题冲冲冲。

 

『C++』

#include <iostream>
#include <vector>
using namespace std;
#define INF 0x7fffffff

int FieldsN, dist[3000];

struct Edge
{
    int s, e;
    int t;
    Edge() {}
    Edge(int _s, int _e, int _t) :
        s(_s), e(_e), t(_t) {}
};

vector<Edge> edges;

bool Bellmen_Ford(int v)
{
    int s, e, t;
    int Size = edges.size();

    for (int k = 1; k <= FieldsN; k++)
        dist[k] = INF;
    dist[v] = 0;
    for (int k = 1; k < FieldsN; k++) {
        for (int i = 0; i < Size; i++) {
            s = edges[i].s;
            e = edges[i].e;
            t = edges[i].t;
            if (dist[s] != INF && dist[s] + t < dist[e])
                dist[e] = dist[s] + t;
        }
    }
    for (int i = 0; i < Size; i++) {
        s = edges[i].s;
        e = edges[i].e;
        t = edges[i].t;
        if (dist[s] + t < dist[e]) return true;
    }
    return false;
}

int main()
{
    int F, M, W, S, E, T;

    cin >> F;
    while (F--) {
        edges.clear();
        cin >> FieldsN >> M >> W;
        while (M--) {
            cin >> S >> E >> T;
            edges.push_back(Edge(S, E, T));
            edges.push_back(Edge(E, S, T));
        }
        while (W--) {
            cin >> S >> E >> T;
            edges.push_back(Edge(S, E, -T));
        }

        if (Bellmen_Ford(1)) printf("YES
");
        else printf("NO
");
    }

    //system("pause");
    return 0;
}