POJ-3259 Wormholes（判断负环、模板）

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

题目大意：N条双向边，W条单向虫洞，走过每条边花费的时间是正的，走虫洞会让时间倒流，每条边和虫洞的花费已知，问小明从起点开始转一圈再回到起点的时间，能不能是在出发之前。
题目解析：用bellman_ford算法，判断是不是存在负环。

代码如下：

 # include<iostream>
 # include<cstdio>
 # include<cstring>
 # include<algorithm>
 using namespace std;
 const int INF=<<;
 struct edge
 {
     int fr,to,w,nxt;
 };
 edge e[];
 int head[],n,cnt,dis[];
 void add(int u,int v,int w)
 {
     e[cnt].fr=u;
     e[cnt].to=v;
     e[cnt].w=w;
     //e[cnt].nxt=head[u];
     //head[u]=cnt++;
     ++cnt;
 }
 bool bellman_ford()
 {
     int i,j;
     fill(dis,dis+n+,INF);
     dis[]=;
     for(i=;i<n;++i){
         for(j=;j<cnt;++j){
             if(dis[e[j].fr]!=INF&&dis[e[j].to]>dis[e[j].fr]+e[j].w){
                 dis[e[j].to]=dis[e[j].fr]+e[j].w;
                 if(i==n-)
                     return true;
             }
         }
     }
     return false;
 }
 int main()
 {
     int T,s,t;
     scanf("%d",&T);
     int a,b,c;
     while(T--)
     {
         cnt=;
         memset(head,-,sizeof(head));
         scanf("%d%d%d",&n,&s,&t);
         while(s--)
         {
             scanf("%d%d%d",&a,&b,&c);
             add(a,b,c);
             add(b,a,c);
         }
         while(t--)
         {
             scanf("%d%d%d",&a,&b,&c);
             add(a,b,-c);
         }
         if(bellman_ford())
             printf("YES\n");
         else
             printf("NO\n");
     }
     return ;
 }