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Wormholes
Time Limit: 2000MS | Memory Limit: 65536K | |
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Total Submissions: 30169 | Accepted: 10914 |
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
代码语言：javascript复制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 8Sample Output
代码语言：javascript复制NO
YESHint
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
AC代码：
代码语言：javascript复制#include<iostream>using namespace std;struct Point{ int s,e,t;}a[10000];int se;int n,m,w;int bell_man(int start){ int dis[10000]; for(int i=1;i<=n;i ) dis[i]=999999; dis[start]=0; for(int i=1;i<n;i ) for(int j=0;j<se;j ) dis[a[j].e] = dis[a[j].e] > dis[a[j].s] a[j].t ? dis[a[j].s] a[j].t : dis[a[j].e]; for(int i=0;i<se;i ){ if(dis[a[i].e] > dis[a[i].s] a[i].t) return 1; } return 0;}int main(){ int T; cin>>T; while(T--){ se=0; cin>>n>>m>>w; for(int i=0;i<m;i ){ int s,e,t; cin>>s>>e>>t; a[se].s=s; a[se].e=e; a[se ].t=t; a[se].s=e; a[se].e=s; a[se ].t=t; } for(int i=0;i<w;i ){ int s,e,t; cin>>s>>e>>t; a[se].s=s; a[se].e=e; a[se ].t=-t; } //int k; //for(k=1;k<=n;k ){ //事实上正确的起点应该要历遍全部点。可是这种超时了 //这个题目仅仅要1点就能够了。算是题目的一个非常大漏洞吧，数据太水了 if(bell_man(1)){ cout<<"YES"<<endl; //break; } //} //if(k>n) else cout<<"NO"<<endl; } return 0;}发布者：全栈程序员栈长，转载请注明出处：https://javaforall.cn/116729.html原文链接：https://javaforall.cn
