POJ 3259——Wormholes 图论

原题如下

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.

题目大意：一个农夫，在他的农场里有m条双向路径和w个单向虫洞。经过每条路需要耗费t秒时间，穿越每个虫洞可以使时间倒退t秒。求他能否在他出发时间之前到达出发地点？

分析：显然是图论相关问题。一个虫洞就相当于负的边权。要想达到要求，必须得通过某个顶点两次。但是如果不存在负圈，那么最短路最多经过n-1条边，在bellman-ford算法中，就相当于只能更新n-1次，也不可能经过同一个顶点两次。

因此题目也就转化成了，图中是否存在负圈。

代码如下：

#include <stdio.h>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <queue>
#include <ctype.h>
#include <vector>
using namespace std;
const int MAXN=10000;
const int INF=1000000;

struct edge{
    int from,to,cost;
};
edge es[MAXN];
int d[MAXN];
int f,n,m,w;

bool find_negative_loop(){
    memset(d,0,sizeof(d));
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<2*m+w;j++)
        {
            edge e=es[j];
            if(d[e.to]>d[e.from]+e.cost)
            {
                d[e.to]=d[e.from]+e.cost;
                if(i==n-1)
                    return true;
            }
        }
    }
    return false;
}
int main()
{
    cin>>f;
    while(f--)
    {
        cin>>n>>m>>w;
    int s,e,t;
    for(int i=0;i<2*m;)
    {
        cin>>s>>e>>t;
        es[i].from=s;
        es[i].to=e;
        es[i].cost=t;
        i++;
        es[i].from=e;
        es[i].to=s;
        es[i].cost=t;
        i++;
    }
    for(int i=2*m;i<w+2*m;i++)
    {
        cin>>s>>e>>t;
        es[i].from=s;
        es[i].to=e;
        es[i].cost=-t;
    }
    if(find_negative_loop())
        printf("YES\n");
    else
        printf("NO\n");
    }
}