poj 3259(bellman最短路径)[通俗易懂]

poj 3259(bellman最短路径)

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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..NM (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

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++){          //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;}

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