poj3259 Wormholes(最短路)

Wormholes Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 48077 Accepted: 17756

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, andW (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 toF (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 这次我认识到,在循环里调用函数会花费大量的时间。。。 这道题最好使用bellman_ford复杂度低。。。。 #include<stdio.h> #define N 505 d[N][N]; int n,m,k; int min(int x,int y) { return x>y?y:x; } int Floyd(void) { int i,j,k,t; for(k=1;k<=n;k++) for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { t=d[i][k]+d[k][j]; if(d[i][j]>t)d[i][j]=t; //如果使用d[i][j]=min(d[i][j],d[i][k]+d[k][j]);会超时 } if(d[i][i]<0) return 1; } return 0; } int main(void) { int t; scanf("%d",&t); while(t--) { int a,b,c,i; scanf("%d%d%d",&n,&m,&k); memset(d,0x3f,sizeof(d)); for(i=1;i<=n;i++) d[i][i]=0; for(i=1;i<=m;i++) { scanf("%d%d%d",&a,&b,&c); if(c<d[a][b]) d[a][b]=d[b][a]=c; } for(i=1;i<=k;i++) { scanf("%d%d%d",&a,&b,&c); d[a][b]=-c; } if(!Floyd()) printf("NO\n"); else printf("YES\n"); } }