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poj 3259(bellman最短路径)

2024-07-13 03:10:16   227人阅读

Wormholes
Time Limit: 2000MS Memory Limit: 65536K
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

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

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

AC代码：

#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++){          //其实正确的起点应该要历遍所有点，但是这样的超时了
                                     //这个题目只要1点就可以了，算是题目的一个很大漏洞吧，数据太水了
            if(bell_man(1)){
                cout<<"YES"<<endl;
                //break;
            }
        //}
        //if(k>n)
        else
            cout<<"NO"<<endl;
    }
    return 0;
}



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