POJ 3259 Wormholes(Bellman-Ford)

题目网址：http://poj.org/problem?id=3259

题目：

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 52198		Accepted: 19426

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.

 

思路：

我们根本不需要关心他所处的起点的具体位置，我们只需要判断是否有负权环即可，所以将所有的dist[i]都初始化为无穷大。有负权环的话就输出YES，没有的话就输出NO。很自然地就会想到Bellman-Ford算法。判断第n次循环，是否还会松弛，如果还需要就说明有负权环。 这道题需要注意的一点是：虫洞是单向边，路径是双向边。

 

代码：

 #include <cstdio>
 #include <iostream>
 #include <vector>
 #include <algorithm>
 using namespace std;
 const int inf = ;
 struct node{
     int v,u,w;
 };
 vector<node>v;
 int n,m,w;
 int dist[];
 node x;
 bool relax(int j){//松弛操作
     if(dist[v[j].u]>dist[v[j].v]+v[j].w){
         dist[v[j].u]=dist[v[j].v]+v[j].w;
         return true;
     }
     return false;
 }
 bool bellman_ford(){
     for (int i=; i<=n; i++) {
         dist[i]=inf;
     }
     for (int i=; i<n-; i++) {
         int flag=;
         for (int j=; j<v.size(); j++) {
             if(relax(j))    flag=;
         }
         if(!flag)   return false;
     }
     for (int j=; j<v.size(); j++) {//核心
         if(relax(j))    return true;
     }
     return false;
 }
 int main(){
     int t;
     cin>>t;
     while (t--) {
         int ok=;
         v.clear();
         cin>>n>>m>>w;
         for (int i=; i<m; i++) {
             cin>>x.v>>x.u>>x.w;
             v.push_back(x);
             swap(x.v, x.u);
             v.push_back(x);
         }
         for (int i=; i<w; i++) {
             cin>>x.v>>x.u>>x.w;
             x.w=-x.w;
             v.push_back(x);
         }
         if (bellman_ford())    printf("YES\n");
         else printf("NO\n");
     }
     return ;
 }