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.
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.
1 #include<iostream> 2 #include<string> 3 #include<cstring> 4 #include<cmath> 5 #include<cstdio> 6 #include<queue> 7 using namespace std; 8 double dis[10001]; 9 int n,m,u,v; 10 int r; 11 struct edge 12 { 13 int u,v; 14 int r; 15 }e[1000001]; 16 int i,j,k,s; 17 void bellman_ford() 18 { 19 //cout<<"M: "<<m<<endl; 20 for(int it=0;it<=n;it++) 21 dis[it]=0.0; 22 for(int it=1;it<n;it++)//松弛n-1次 23 { 24 int flag=1; 25 for(int jt=0;jt<m;jt++) 26 { 27 int u=e[jt].u; 28 int v=e[jt].v; 29 int r=e[jt].r; 30 31 if(dis[v]>dis[u]+r) 32 { 33 dis[v]=dis[u]+r; 34 flag=0; 35 } 36 //cout<<"dis["<<v<<"]: "<<dis[v]<<endl; 37 } 38 if(flag==1) //找不到要松弛的提前返回 39 { 40 puts("NO"); 41 return; 42 } 43 } 44 for(int it=0;it<m;it++)//找是否存在正环 45 if(dis[e[it].v]>dis[e[it].u]+e[it].r) 46 { 47 puts("YES"); 48 return; 49 } 50 puts("NO"); 51 } 52 int main() 53 { 54 int cas; 55 scanf("%d",&cas); 56 while(cas--) 57 { 58 k=0; 59 scanf("%d%d%d",&n,&m,&s); 60 for(i=1;i<=m;i++) 61 { 62 scanf("%d%d%d",&u,&v,&r); 63 e[k].u=u; 64 e[k].v=v; 65 e[k++].r=r; 66 e[k].u=v; 67 e[k].v=u; 68 e[k++].r=r; 69 } 70 for(i=1;i<=s;i++) 71 { 72 scanf("%d%d%d",&u,&v,&r); 73 e[k].u=u; 74 e[k].v=v; 75 e[k++].r=-r; 76 } 77 m=k; 78 bellman_ford(); 79 } 80 return 0; 81 }
转载于:https://www.cnblogs.com/sdau--codeants/p/3378075.html