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| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 34465 | Accepted: 12585 |
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
Output
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
Hint
1 #include<stdio.h> 2 #include<string.h> 3 4 const int INF=99999999; 5 6 struct Node{ 7 int u; 8 int v; 9 int o; 10 }; 11 12 Node edge[10250]; 13 int dis[550]; 14 int n,m,w,nn; 15 16 void relax(int u,int v,int o) 17 { 18 if(dis[v]>dis[u]+o) 19 dis[v]=dis[u]+o; 20 } 21 22 bool Bellan_Ford() 23 { 24 int i,j,k; 25 for(i=1;i<=n;i++) 26 dis[i]=INF; 27 28 for(i=1;i<=n-1;i++) 29 { 30 for(j=1;j<=nn;j++) 31 relax(edge[j].u,edge[j].v,edge[j].o); 32 } 33 34 for(i=1;i<=nn;i++) 35 { 36 if(dis[edge[i].v]>dis[edge[i].u]+edge[i].o) 37 { 38 return false; 39 } 40 } 41 return true; 42 } 43 44 int main() 45 { 46 int F,i,j,k,flg; 47 int s,e,t; 48 scanf("%d",&F); 49 while(F--) 50 { 51 flg=0;nn=1; 52 scanf("%d %d %d",&n,&m,&w); 53 for(i=1;i<=m;i++) 54 { 55 scanf("%d %d %d",&edge[nn].u,&edge[nn].v,&edge[nn].o); 56 nn++; 57 edge[nn].u=edge[nn-1].v,edge[nn].v=edge[nn-1].u,edge[nn].o=edge[nn-1].o; 58 nn++; 59 } 60 for(i=1;i<=w;i++) 61 { 62 scanf("%d %d %d",&edge[nn].u,&edge[nn].v,&edge[nn].o); 63 edge[nn].o=-edge[nn].o; 64 nn++; 65 } 66 nn--; 67 if(Bellan_Ford()==false) 68 { 69 flg=1; 70 } 71 if(flg==1) 72 printf("YES\n"); 73 else 74 printf("NO\n"); 75 } 76 return 0; 77 }
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原文地址：http://www.cnblogs.com/cyd308/p/4545043.html
