zoukankan      html  css  js  c++  java
  • HDU 2680 Choose the best route

    Choose the best route

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5783    Accepted Submission(s): 1866

    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 5
    1 2 2
    1 5 3
    1 3 4
    2 4 7
    2 5 6
    2 3 5
    3 5 1
    4 5 1
    2
    2 3
    4 3 4
    1 2 3
    1 3 4
    2 3 2
    1
    1
     
    Sample Output
    1
    -1
     
    Author
    dandelion
     
    Source
     
    Recommend
    lcy   |   We have carefully selected several similar problems for you:  2112 1874 1217 1142 1385 
     
    思路:dijkstra
     
    代码:
    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <cmath>
    #include <cstdlib>
    #include <algorithm>
    using namespace std;
    const int MAXN = 100000000;
    int map[1010][1010];
    int dist[1010];
    int n,m,end;
    int w;
    int first;
    int the_last_flag;
    int a,b,tim;
    void Dijkstra(int n,int dist[1010],int map[1010][1010],int s)
    {
        int minn;
        bool p[1010];
        for(int i = 0;i <= n;i ++)
        {
            if(i != s)
            {
                dist[i] = map[s][i];
                p[i] = false;
            }
        }
        p[s] = true;
        dist[s] = 0;
        for(int i = 1;i <= n;i ++)
        {
            the_last_flag = 0;
            minn = MAXN;
            for(int j = 1;j <= n;j ++)
            {
                if(!p[j] && dist[j] < minn)
                {
                    minn = dist[j];
                    the_last_flag = j;
                }
            }
            if(the_last_flag == 0)
                break ;
            if(the_last_flag == end)
                break ;
            p[the_last_flag] = true;
            for(int j = 1;j <= n;j ++)
            {
                if(!p[j] && map[the_last_flag][j] != MAXN &&
                   dist[j] > dist[the_last_flag] + map[the_last_flag][j])
                   {
                       dist[j] = dist[the_last_flag] + map[the_last_flag][j];
                   }
            }
        }
    }
    int main()
    {
        while(~scanf("%d%d%d",&n,&m,&end))
        {
            for(int i = 0;i <= n;i ++)
                for(int j = 0;j <= n;j ++)
                {
                    if(i == j)
                       map[i][j] = 0;
                    else
                       map[i][j] = MAXN;
                }
            while(m --)
            {
                scanf("%d%d%d",&a,&b,&tim);
                if(map[a][b] > tim)
                   map[a][b] = tim;
            }
            scanf("%d",&w);
            while(w --)
            {
                scanf("%d",&first);
                map[0][first] = 0;
            }
            Dijkstra(n,dist,map,0);
            if(dist[end] == MAXN)
                 printf("-1
    ");
            else
                 printf("%d
    ",dist[end]);
        }
        return 0;
    }
  • 相关阅读:
    MyEclipse安装Freemarker插件
    C# winform webbrowser如何指定内核
    tfs清除服务器印射
    在sql中怎样把int行转化成我想要的格式,比如把1转化为'001',
    为项目加入npoi
    .net邮件错误 :The specified string is not in the form required for a subject.
    C#编程总结(十)字符转码
    ,字符串,列表,元组,字典
    简单登录,编码解码,
    密码*** ,continue,等差求和
  • 原文地址:https://www.cnblogs.com/GODLIKEING/p/3407406.html
Copyright © 2011-2022 走看看