Wormholes
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 32474 Accepted: 11808
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
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
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
USACO 2006 December Gold
/*Source Code Problem: 3259 User: Grant Yuan Memory: 224K Time: 125MS Language: C++ Result: Accepted*/ Source Code #include<iostream> #include<algorithm> #include<cstdio> #include<cstring> using namespace std; int n,m,w; struct edges { int from,to,cost; }; int d[507]; edges edge[6000]; int all=0; bool Bellman_Ford() { bool flag; memset(d,0,sizeof(d)); for(int i=1;i<=n;i++) { flag=false; for(int k=0;k<all;k++) { if(d[edge[k].to]>d[edge[k].from]+edge[k].cost){ flag=true; d[edge[k].to]=d[edge[k].from]+edge[k].cost; } } if(i==n&&flag) return true; } return false; } int main() { // freopen("in.txt","r",stdin); int f;int a,b,c; scanf("%d",&f); while(f--){ memset(edge,0,sizeof(edge)); all=0; scanf("%d%d%d",&n,&m,&w); for(int i=0;i<m;i++) { scanf("%d%d%d",&a,&b,&c); edge[all].from=a;edge[all].to=b;edge[all++].cost=c; edge[all].from=b;edge[all].to=a;edge[all++].cost=c; } for(int i=0;i<w;i++) { scanf("%d%d%d",&a,&b,&c); edge[all].from=a;edge[all].to=b;edge[all++].cost=-c; } bool ans=Bellman_Ford(); if(ans) printf("YES "); else printf("NO "); } return 0; }