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; }