Time Limit: 2000MS | Memory Limit: 65536K | |
tal Submissions: 11389 | Accepted: 3998 |
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 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
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 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.
bellman-ford
#include <iostream> #include <cstring> #include <cstdio> using namespace std; typedef struct { int s,e; int t; }edge; edge biway[5001]; edge oneway[201]; int d[501]; int F; int main() { int N,M,W; int s,e,t; scanf("%d",&F); for(int f=1; f<=F; f++) { memset(biway,0,sizeof(biway)); memset(oneway,0,sizeof(oneway)); memset(d,10000,sizeof(d)); scanf("%d%d%d",&N,&M,&W); for(int i=1; i<=M; i++) { scanf("%d%d%d",&s,&e,&t); biway[i].s=s; biway[i].e=e; biway[i].t=t; biway[i+2500].s=e; biway[i+2500].e=s; biway[i+2500].t=t; } for(int i=1; i<=W; i++) { scanf("%d%d%d",&s,&e,&t); oneway[i].s=s; oneway[i].e=e; oneway[i].t=-t; } //bellman-ford for(int i=0; i<N; i++) { for(int j=1; j<=M; j++) { //relax if( d[biway[j].e] > d[biway[j].s] + biway[j].t) d[biway[j].e] = d[biway[j].s] + biway[j].t; if( d[biway[j+2500].e] > d[biway[j+2500].s] + biway[j+2500].t) d[biway[j+2500].e] = d[biway[j+2500].s] + biway[j+2500].t; } for(int j=1; j<=W; j++) { if(d[oneway[j].e] > d[oneway[j].s]+oneway[j].t) d[oneway[j].e]=d[oneway[j].s]+oneway[j].t; } } int isRelax=0; for(int j=1; j<=M; j++) { //relax if( d[biway[j].e] > d[biway[j].s] + biway[j].t) isRelax=1; if( d[biway[j+2500].e] > d[biway[j+2500].s] + biway[j+2500].t) isRelax=1; } for(int j=1; j<=W; j++) { if(d[oneway[j].e] > d[oneway[j].s]+oneway[j].t) isRelax=1; } if(isRelax) printf("YES\n"); else printf("NO\n"); } return 0; }