Wormholes-------最短路径Bellman ford

最新推荐文章于 2021-08-05 17:22:25 发布

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这篇博客探讨了农民John如何利用虫洞在他的农场中进行时间旅行。每个农场由多个字段、路径和虫洞组成。John希望从某个字段出发，经过一些路径和虫洞，回到起点，时间早于出发时间。博客通过输入描述农场的连接，并展示了一个Bellman-Ford算法的实现，用于检查这个目标是否可能。如果可能，输出'YES'，否则输出'NO'。

Wormholes

时间限制(普通/Java):3000MS/10000MS 运行内存限制:65536KByte
总提交:60 测试通过:26

描述

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.

输入

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.

输出

Lines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).

样例输入

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

样例输出

NO
YES

#include<iostream> using namespace std; typedef struct { int s; int d; int w; }Node; Node p[5204],e; int t,n,m,w,i,sum,temp; void BellmanFord() { int i,j,d[1000]={0}; bool flag; for(i=1;i<=n;i++) { flag=true; for(j=1;j<=sum;j++) { if(d[p[j].d]>p[j].w+d[p[j].s]) { d[p[j].d]=p[j].w+d[p[j].s]; flag=false; } } if(flag==true) { cout<<"NO"<<endl; return; } if(flag==false&&i==n) { cout<<"YES"<<endl; return; } } } int main() { cin>>t; while(t--) { sum=0; cin>>n>>m>>w; for(i=1;i<=m;i++) { cin>>e.s>>e.d>>e.w; p[++sum]=e; temp=e.s; e.s=e.d; e.d=temp; p[++sum]=e; } for(i=1;i<=w;i++) { cin>>e.s>>e.d>>e.w; e.w=-e.w; p[++sum]=e; } BellmanFord(); } return 0; }