POJ 3259 Wormholes 最短路径

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 48298 Accepted: 17818

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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

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

#include <stdio.h>#include <iostream>#include <algorithm>#define INF 0x3f3f3f3 using namespace std;int f,n,m,w,map[505][505];int fly(){int flag=0,i,j,k,tmp;for(k=1;k<=n;k++){for(i=1;i<=n;i++){for(j=1;j<=n;j++){tmp=map[i][k]+map[k][j];if(map[i][j] > tmp)map[i][j]=tmp;}if(map[i][i]<0)return 1;}}return 0;}int main(){int k,i,j,p,q,t,flag;scanf("%d",&f);while(f--){scanf("%d%d%d",&n,&m,&w);for(i=0;i<=n;i++){for(j=0;j<=n;j++){map[i][j]=INF;}}flag=0;for(i=0;i<m;i++){//pathscanf("%d%d%d",&p,&q,&t);if(t < map[p][q])map[p][q]=map[q][p]=t;}for(i=0;i<w;i++){//wormholesscanf("%d%d%d",&p,&q,&t);map[p][q]=(-1)*t;}flag=fly();if(flag)printf("YES\n");elseprintf("NO\n");}return 0;}