Wormholes （弗洛伊德 ）

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 toF (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, <i>F</i>. <i>F</i> farm descriptions follow. <br>Line 1 of each farm: Three space-separated integers respectively: <i>N</i>, <i>M</i>, and <i>W</i> <br>Lines 2..<i>M</i>+1 of each farm: Three space-separated numbers (<i>S</i>, <i>E</i>, <i>T</i>) that describe, respectively: a bidirectional path between <i>S</i> and <i>E</i> that requires <i>T</i> seconds to traverse. Two fields might be connected by more than one path. <br>Lines <i>M</i>+2..<i>M</i>+<i>W</i>+1 of each farm: Three space-separated numbers (<i>S</i>, <i>E</i>, <i>T</i>) that describe, respectively: A one way path from <i>S</i> to <i>E</i> that also moves the traveler back <i>T</i> seconds.

Output

Lines 1..<i>F</i>: 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

题目大概：

几个农场之间，有双向通道（正权值），有单向通道（负权值），找出其中是否有负权回路。

思路：

用弗洛伊德算法。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

int n,m,h;
int dis[6000];
int s=0;
struct point
{
    int l,r,shu;
}a[6000];
int fin()
{
    dis[1]=0;
    for(int i=1;i<n-1;i++)
    {
        for(int j=1;j<=s;j++)
        {
            if(dis[a[j].r]>dis[a[j].l]+a[j].shu)
            dis[a[j].r]=dis[a[j].l]+a[j].shu;
        }
    }
    for(int j=1;j<=s;j++)
    if(dis[a[j].r]>dis[a[j].l]+a[j].shu)return 0;

    return 1;


}


int main()
{
    int t;
    cin>>t;
    for(int i=1;i<=t;i++)
    {    memset(dis,10003,sizeof(dis));
    memset(a,0,sizeof(a));
         s=0;
         cin>>n>>m>>h;

         for(int j=1;j<=m;j++)
         {int z1,z2,z3;
          cin>>z1>>z2>>z3;
         a[++s].l=z1;
         a[s].r=z2;
         a[s].shu=z3;
         a[++s].l=z2;
         a[s].r=z1;
         a[s].shu=z3;

        }
         for(int j=1;j<=h;j++)
         {int z1,z2,z3;
          cin>>z1>>z2>>z3;
          a[++s].l=z1;
         a[s].r=z2;
         a[s].shu=-z3;
         }


         int pp=fin();
         if(pp==1)cout<<"NO"<<endl;
         else cout<<"YES"<<endl;



    }

    return 0;
}