Wormholes

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 81   Accepted Submission(s) : 27

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  

Source PKU  

题意：

农民约翰在农场散步的时候发现农场有大量的虫洞，这些虫洞是非常特别的因为它们都是单向通道，为了方便现在把约翰的农田划分成N快区域，M条道路，W的虫洞。

约翰是时空旅行的粉丝，他希望这样做，在一个区域开始，经过一些道路和虫洞然后回到他原来所在的位置，这样也许他就能见到他自己了。穿越虫洞会回到以前。

思路：其实给出了坐标，这个时候就可以构成一张图，然后将回到从前理解为是否会出现负权环，用bellman-ford就可以解出了

代码：

#include <iostream>
#include <cstring>
#include <string>
#include <stdio.h>
#define inf 0x3f3f3f3f
using namespace std;
struct paths{
int s,e,t;
}p[5210];
 bool flag;
int n,m,w,cnt;
int dis[520];
int main()
{
    int i,j,k,temp;
    int t;
    cin>>t;
    while(t--)
    {
        cin>>n>>m>>w;
        for(i=1;i<=m;i++){
            cin>>p[i].s>>p[i].e>>p[i].t;
            p[i+m].s=p[i].e;p[i+m].e=p[i].s;p[i+m].t=p[i].t; //输入每点情况
        }
        k=2*m+w; //注意这是边数。
        for(i=2*m+1;i<=k;i++)
        {
            cin>>p[i].s>>p[i].e>>temp;
            p[i].t=-temp;
        }
        m=k;

        for(j=1;j<=n;j++) dis[j]=inf;//初始化
        dis[1]=0;
        for(j=1;j<=n;j++)
        {
            flag=0;
            for(k=1;k<=m;k++)
            {
                if(dis[p[k].s]>p[k].t+dis[p[k].e]){

                    dis[p[k].s]=p[k].t+dis[p[k].e];flag=1;
                }
            }

            if(!flag)break;  //优化
        }
        flag=0;
        for(k=1;  k<=m; k++)  //如果仍有大于它的则出现负环
        {
            if(dis[p[k].s]>p[k].t+dis[p[k].e]){flag=1;break;}
        }
        if(flag)cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    return 0;
}