zoukankan      html  css  js  c++  java
  • HDU 1102 Constructing Roads Prim

    Constructing Roads

    Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
    Total Submission(s) : 12   Accepted Submission(s) : 2

    Font: Times New Roman | Verdana | Georgia

    Font Size:

    Problem Description

    There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.
    We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

    Input

    The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.
    Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

    Output

    You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

    Sample Input

    3
    0 990 692
    990 0 179
    692 179 0
    1
    1 2
    

    Sample Output

    179
    

    Source

    kicc
     
    思路:Prim OK 了
     
    代码:
    #include <iostream>
    #include <cstring>
    #include <cstdlib>
    #include <cstdio>
    #include <cmath>
    #include <algorithm>
    using namespace std;
    const int MAXN = 1000000;
    int n,q;
    int a,b;
    int how_long;
    int map[310][310];
    int pre[310];
    bool p[310];
    int dist[310];
    int the_last_flag;
    int the_last_sum;
    void Prim(int n,int dist[310],int map[310][310],int pre[310])
    {
        int minn;
        for(int i = 2;i <= n;i ++)
        {
            p[i] = false;
            dist[i] = map[1][i];
            pre[i] = 1;
        }
        dist[1] = 0;
        p[1] = true;
        for(int i = 1;i <= n - 1;i ++)
        {
            minn = MAXN;
            the_last_flag = 0;
            for(int j = 1;j <= n;j ++)
            {
                if(!p[j] && minn > dist[j])
                {
                    minn = dist[j];
                    the_last_flag = j;
                }
            }
            if(the_last_flag == 0)
               return ;
            p[the_last_flag] = true;
            for(int j = 1;j <= n;j ++)
            {
                if(!p[j] && map[the_last_flag][j] != MAXN && dist[j] > map[the_last_flag][j])
                {
                    dist[j] = map[the_last_flag][j];
                    pre[j] = the_last_flag;
                }
            }
    
        }
    }
    int main()
    {
        while(~scanf("%d",&n))
        {
            for(int i = 1;i <= n;i ++)
               for(int j = 1;j <= n;j ++)
               {
                   if(i == j)
                     map[i][j] = 0;
                   else
                   {
                       map[i][j] = MAXN;
                   }
               }
            for(int i = 1;i <= n;i ++)
               for(int j = 1;j <= n;j ++)
               {
                   scanf("%d",&how_long);
                   map[i][j] = how_long;
               }
            scanf("%d",&q);
            while(q --)
            {
                scanf("%d%d",&a,&b);
                map[a][b] = 0;
                map[b][a] = 0;
            }
            the_last_flag = 0;
            the_last_sum = 0;
            Prim(n,dist,map,pre);
            for(int i = 1;i <= n;i ++)
               the_last_sum += dist[i];
            printf("%d
    ",the_last_sum);
        }
        return 0;
    }
  • 相关阅读:
    Android按返回键退出程序但不销毁,程序后台运行,同QQ退出处理方式
    android 下动态获取控件的id
    Android大图片裁剪终极解决方案 原理分析
    如何使用Android MediaStore裁剪大图片
    最新的Android Sdk 使用Ant多渠道批量打包
    nodejs学习(1)
    C#——企业微信一般操作之一
    html(1)——转圈等待效果+鼠标移动悬浮显示相关信息
    SQL注入小结
    Java实现二叉树地遍历、求深度和叶子结点的个数
  • 原文地址:https://www.cnblogs.com/GODLIKEING/p/3376709.html
Copyright © 2011-2022 走看看