poj3259 bellman——ford Wormholes解绝负权问题

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 35103		Accepted: 12805

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

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.

Source

题目大意：虫洞问题，现在有n个点，m条边，代表现在可以走的通路，比如从a到b和从b到a需要花费c时间，现在在地上出现了w个虫洞，虫洞的意义就是你从a到b话费的时间是-c(时间倒流,并且虫洞是单向的)，现在问你从某个点开始走，能回到从前

解题思路：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

#include<stdio.h>
#include<string.h>
#include<stack>
#include<iostream>
#include<algorithm>
using namespace std;
struct node{
  int u,v,w;
}que[5400];
int n,m,wh;
int Count;
int inf=999999999;
int dis[5000];
bool bellman_ford(){
   memset(dis,inf,sizeof(dis));
   dis[1]=0;
   int flag;
   int a,b,c;
   for(int i=1;i<n;i++){
       flag=0;
       for(int j=0;j<Count;j++){
         a=que[j].u,b=que[j].v,c=que[j].w;
          if(dis[b]>dis[a]+c){
            dis[b]=dis[a]+c;
            flag=1;
          }
       }
       if(!flag)
        break;
   }
   for(int j=0;j<Count;j++){
          a=que[j].u,b=que[j].v,c=que[j].w;
    if(dis[b]>dis[a]+c)
        return true;}
   return false;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
            Count=0;
        scanf("%d%d%d",&n,&m,&wh);
         int t1,t2,t3;
        for(int i=1;i<=m;i++){

            scanf("%d%d%d",&t1,&t2,&t3);
             que[Count].u=t1;
             que[Count].v=t2;
             que[Count].w=t3;
             Count++;
             que[Count].u=t2;
             que[Count].v=t1;
             que[Count].w=t3;
             Count++;
        }
        for(int i=m+1;i<=m+wh;i++){
           scanf("%d%d%d",&t1,&t2,&t3);
           que[Count].u=t1;
           que[Count].v=t2;
           que[Count].w=-t3;
           Count++;
        }
        if(bellman_ford())
            printf("YES
");
        else
            printf("NO
");
    }
    return 0;
}