POJ3259 Wormholes Floyd

Wormholes Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 48494 Accepted: 17878

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 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

123

虫洞问题 ;

 从农场出发 然后 经过虫洞 虫洞   用负权   套用Floyd 模板 即可

/* floyd 最短路径 从某地出发 回到某地 */ #include<stdio.h> #define inf 1<<29 #define N 1000 int map[N][N]; int n,m,w; bool Floyd() { int i,j,k; for(k=1;k<=n;k++) for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { if(map[i][j]>map[i][k]+map[k][j]) map[i][j]=map[i][k]+map[k][j]; } // printf("%d\n",map[i][i]); if(map[i][i]<0) { return true; } } return false; } int main() { int i,j,k; int T; int x,y,z; while(~scanf("%d",&T)) { while(T--) { scanf("%d %d %d",&n,&m,&w); for(i=1;i<=n;i++) for(j=1;j<=n;j++) { if(i==j)map[i][j]=0; else map[i][j]=inf; } for(i=1;i<=m;i++) { scanf("%d %d %d",&x,&y,&z); if(z<map[x][y]) map[x][y]=map[y][x]=z; } for(i=1;i<=w;i++) { scanf("%d %d %d",&x,&y,&z); map[x][y]=-z;//负权值 } // for(i=1;i<=10;i++) // { // for(j=1;j<=10;j++) // printf("%d ",map[i][j]); // printf("\n"); // } printf(Floyd()?"YES\n":"NO\n"); } } return 0; }