zoukankan      html  css  js  c++  java
  • poj 1160 Post Office (间隔DP)

    Post Office
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 15966   Accepted: 8671

    Description

    There is a straight highway with villages alongside the highway. The highway is represented as an integer axis, and the position of each village is identified with a single integer coordinate. There are no two villages in the same position. The distance between two positions is the absolute value of the difference of their integer coordinates. 

    Post offices will be built in some, but not necessarily all of the villages. A village and the post office in it have the same position. For building the post offices, their positions should be chosen so that the total sum of all distances between each village and its nearest post office is minimum. 

    You are to write a program which, given the positions of the villages and the number of post offices, computes the least possible sum of all distances between each village and its nearest post office. 

    Input

    Your program is to read from standard input. The first line contains two integers: the first is the number of villages V, 1 <= V <= 300, and the second is the number of post offices P, 1 <= P <= 30, P <= V. The second line contains V integers in increasing order. These V integers are the positions of the villages. For each position X it holds that 1 <= X <= 10000.

    Output

    The first line contains one integer S, which is the sum of all distances between each village and its nearest post office.

    Sample Input

    10 5
    1 2 3 6 7 9 11 22 44 50

    Sample Output

    9


    题意:在v个村庄中建立p个邮局。求全部村庄到它近期的邮局的距离和。村庄在一条直线上。邮局建在村庄上。

    思路:首先求出在连续的几个村庄上建立一个邮局的最短距离,用数组dis[i][j]表示在第i个村庄和第j个村庄之间建一个邮局的最短距。

    dis[i][j]=dis[i][j-1]+x[j]-x[(i+j)/2]; (村庄位置为x[i])

    用数组dp[i][j]表示在前i个村庄中建立j个邮局的最小距离。即在前k(k<i)个村庄建立j-1个邮局,在k+1到j个村庄建立一个邮局。

    dp[i][j]=min(dp[i][j],dp[k][j-1]+dis[k+1][i]) 


    #include<stdio.h>
    #include<math.h>
    #include<string.h>
    #include<stdlib.h>
    #include<algorithm>
    using namespace std;
    #define N 305
    const int inf=0x3fffffff;
    int dp[N][35]; //在前i个村庄中建立j个邮局的最小耗费
    int dis[N][N];//dis[i][j]:第i个村庄到第j个村庄建一个邮局的最短距离
    int x[N]; //村庄位置
    int main()
    {
        int v,p,i,j,k;
        while(scanf("%d%d",&v,&p)!=-1)
        {
            for(i=1;i<=v;i++)
                scanf("%d",&x[i]);
            //memset(dis,0,sizeof(dis));
            for(i=1;i<=v;i++)
            {
                for(j=i+1;j<=v;j++)
                {
                    dis[i][j]=dis[i][j-1]+x[j]-x[(i+j)/2];
                }
            }
            for(i=1;i<=v;i++)
            {
                dp[i][i]=0; //一个村庄一个邮局距离为零
                dp[i][1]=dis[1][i]; //前i个村庄建立一个邮局
            }
            for(j=2;j<=p;j++)
            {
                for(i=j+1;i<=v;i++)
                {
                    dp[i][j]=inf;
                    for(k=j-1;k<i;k++)
                    {
                        dp[i][j]=min(dp[i][j],dp[k][j-1]+dis[k+1][i]);
                    }
                }
            }
            printf("%d
    ",dp[v][p]);
        }
        return 0;
    }
    

    
    

    版权声明:本文博客原创文章。博客,未经同意,不得转载。

  • 相关阅读:
    .net序列化和反序列化(一)——自动序列化
    在Sql Server 2005使用公用表表达式CTE简化复杂的查询语句
    使用JQuery与iframe交互
    FCKeditor自定义工具栏和定义多个工具栏
    FCKeditor自定义非空验证
    PHP5.3.6的IIS配置
    Linux下缓存服务器的应用
    PHP采集程序中常用的函数
    关于PHP5.3.x和Zend Optimizer(Zend Guard Loader),以及shopex4.8.5安装的问题
    SQLserver数据库还原出现错误112(磁盘空间不足)的解决办法
  • 原文地址:https://www.cnblogs.com/bhlsheji/p/4739544.html
Copyright © 2011-2022 走看看