Poj3259(floyd)

POJ3259 - Wormholes

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 Tseconds 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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

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 accompl

 

     题意其实就是找是否存在负权边，故不能用迪杰斯特拉，Bellman 或者SPFA都可以 所以我敲了最简单的floyd。。。

#include<cstdio>  #include<cstring>  #include<algorithm>  #include<iostream>  using namespace std;  int map[505][505],n,m,k,num=0;  int floyd()  {      int i,j,k,f=0;      for(k=1;k<=n;k++)          for(i=1;i<=n;i++){              for(j=1;j<=n;j++)              {                  int t=map[i][k]+map[k][j];                  if(map[i][j]>t)map[i][j]=t;              }              if(map[i][i]<0)return 1;      }      return f;  }  int main()  {      int t;      scanf("%d",&t);      while(t--)      {          int i,j,a,b,c;          scanf("%d%d%d",&n,&m,&k);          memset(map,0x3f3f3f3f,sizeof(map));          for(i=1;i<=n;i++)map[i][i]=0;          for(i=1;i<=m;i++)          {              scanf("%d%d%d",&a,&b,&c);              if(c<map[a][b])map[a][b]=map[b][a]=c;          }          for(i=1;i<=k;i++)          {              scanf("%d%d%d",&a,&b,&c);              map[a][b]=-c;          }          if(floyd())printf("NO\n");          else printf("YES\n");      }      return 0;  }