poj-3259-Wormholes

Wormholes Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 48286 Accepted: 17817 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.

思路：给出共有n个农场，m条路径（双向），w个虫洞（单向）；其中虫洞是时间倒流，即负权值。就是求出有无负值，若有，则可以回到过去看到自己，输出YES，反之，输出NO。

#include <stdio.h> #include <string.h> #define inf 10010 struct Node{ int u,v,l; }e[2700];//存储u到v路程 int f,n,m,w; int bellman(){ int i,j,flag,cnt; int d[2700]; flag=1;cnt=1; for(i=1;i<=n;i++){ d[i]=inf; } while(flag){ flag=0; if(cnt++>n) return 1; for(i=1;i<=m;i++){ if(d[e[i].u]+e[i].l < d[e[i].v]){ d[e[i].v] = d[e[i].u]+e[i].l; flag=1; } if(d[e[i].v]+e[i].l < d[e[i].u]){ d[e[i].u] = d[e[i].v]+e[i].l; flag=1; } } //2--m+1行双向路径 for(;i<=m+w;i++){ if(d[e[i].v] > d[e[i].u]-e[i].l){ d[e[i].v] = d[e[i].u]-e[i].l; flag=1; } }//m+2--m+w+1行单向即负权值，所以减去 } return 0; } int main() { int i; scanf("%d",&f);//f次测试数据 while(f--){ scanf("%d%d%d",&n,&m,&w); for(i=1;i<=m+w;i++){ scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].l); } if(bellman()){ printf("YES\n"); } else printf("NO\n"); } return 0; }