Wormholes 和上题一样，不过这道题 求负环，代码稍微变一下就行

Problem 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

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bellman_ford算法求负环

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#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
using namespace std;
double dis[10001];
int n,m,u,v;
int r;
struct edge
{
    int u,v;
    int r;
}e[1000001];
int i,j,k,s;
void bellman_ford()
{
    //cout<<"M: "<<m<<endl;
    for(int it=0;it<=n;it++)
      dis[it]=0.0;
    for(int it=1;it<n;it++)//松弛n-1次
    {
        int flag=1;
        for(int jt=0;jt<m;jt++)
        {
            int u=e[jt].u;
            int v=e[jt].v;
            int r=e[jt].r;

          if(dis[v]>dis[u]+r)
          {
            dis[v]=dis[u]+r;
            flag=0;
          }
          //cout<<"dis["<<v<<"]: "<<dis[v]<<endl;
        }
        if(flag==1)  //找不到要松弛的提前返回
        {
            puts("NO");
            return;
        }
    }
    for(int it=0;it<m;it++)//找是否存在正环
      if(dis[e[it].v]>dis[e[it].u]+e[it].r)
       {
           puts("YES");
           return;
       }
    puts("NO");
}
int main()
{
    int cas;
    scanf("%d",&cas);
    while(cas--)
    {
      k=0;
      scanf("%d%d%d",&n,&m,&s);
      for(i=1;i<=m;i++)
      {
          scanf("%d%d%d",&u,&v,&r);
          e[k].u=u;
          e[k].v=v;
          e[k++].r=r;
          e[k].u=v;
          e[k].v=u;
          e[k++].r=r;
      }
      for(i=1;i<=s;i++)
      {
          scanf("%d%d%d",&u,&v,&r);
          e[k].u=u;
          e[k].v=v;
          e[k++].r=-r;
      }
      m=k;
      bellman_ford();
    }
    return 0;
}