HDOJ Choose the best route 2680【最短路Dijkstra+反向建图】

Choose the best route

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10257    Accepted Submission(s): 3310

Problem Description

One day , Kiki wants to visit one of her friends. As she is liable to carsickness , she wants to arrive at her friend’s home as soon as possible . Now give you a map of the city’s traffic route, and the stations which are near Kiki’s home so that she can take. You may suppose Kiki can change the bus at any station. Please find out the least time Kiki needs to spend. To make it easy, if the city have n bus stations ,the stations will been expressed as an integer 1,2,3…n.

 

Input

There are several test cases. 
Each case begins with three integers n, m and s,(n<1000,m<20000,1=<s<=n) n stands for the number of bus stations in this city and m stands for the number of directed ways between bus stations .(Maybe there are several ways between two bus stations .) s stands for the bus station that near Kiki’s friend’s home.
Then follow m lines ,each line contains three integers p , q , t (0<t<=1000). means from station p to station q there is a way and it will costs t minutes .
Then a line with an integer w(0<w<n), means the number of stations Kiki can take at the beginning. Then follows w integers stands for these stations.

 

Output

The output contains one line for each data set : the least time Kiki needs to spend ,if it’s impossible to find such a route ,just output “-1”.

 

Sample Input

5 8 51 2 21 5 31 3 42 4 72 5 62 3 53 5 14 5 122 34 3 41 2 31 3 42 3 211

 

Sample Output

1-1

 

Author

dandelion

 

Source

2009浙江大学计算机研考复试（机试部分）——全真模拟

 

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题意：主角晕车，要去朋友家，要最快到达。。。n个站，m个公交车，有向图，终点s。给你w个起点，代表可以从w开始出发。求最短到达s的时间。

题目分析：最短路。反向考虑。因为w较多，如果每个循环里都放一个dijkstra判断最短路，那么一定会超时。所以，可以考虑反向建边，把终点起点的位置互换。

#include <stdio.h>#include <math.h>#include <vector>#include <queue>#include <string>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>#define INF 0x3f3f3f3fusing namespace std;const int MAXN = 1010;bool used[MAXN];int d[MAXN];int cost[MAXN][MAXN];int N;void Dijkstra(int s){    for(int i=1;i<=N;i++){        d[i]=INF;        used[i]=false;    }    d[s]=0;    while(true){        int v=-1;        for(int i=1;i<=N;i++)            if(!used[i]&&(v==-1||d[v]>d[i])) v=i;        if(v==-1) break;        used[v]=true;        for(int i=1;i<=N;i++)            d[i]=min(d[i],d[v]+cost[v][i]);    }}int main(){    int m,s;    while(scanf("%d%d%d",&N,&m,&s)!=EOF){        for(int i=1;i<=N;i++)            for(int j=1;j<=N;j++){                if(i==j) cost[i][j]=0;                else cost[i][j]=INF;            }        int a,b,c;        for(int i=0;i<m;i++){            scanf("%d%d%d",&a,&b,&c);            if(cost[b][a]>c)                cost[b][a]=c;        }        int w,g;        scanf("%d",&w);        int minn=INF;        Dijkstra(s);        while(w--){            scanf("%d",&g);            minn=min(minn,d[g]);        }        if(minn==INF) printf("-1\n");        else printf("%d\n",minn);    }    return 0;}