HDU_1227_Fast Food_动态规划

时间：2016-05-07 18:10:17 阅读：151 评论：0 收藏：0 [点我收藏+]

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链接：http://acm.hdu.edu.cn/showproblem.php?pid=1227

Fast Food

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2695 Accepted Submission(s): 1142

Problem Description

The fastfood chain McBurger owns several restaurants along a highway. Recently, they have decided to build several depots along the highway, each one located at a restaurant and supplying several of the restaurants with the needed ingredients. Naturally, these depots should be placed so that the average distance between a restaurant and its assigned depot is minimized. You are to write a program that computes the optimal positions and assignments of the depots.

To make this more precise, the management of McBurger has issued the following specification: You will be given the positions of n restaurants along the highway as n integers d1 < d2 < ... < dn (these are the distances measured from the company‘s headquarter, which happens to be at the same highway). Furthermore, a number k (k <= n) will be given, the number of depots to be built.

The k depots will be built at the locations of k different restaurants. Each restaurant will be assigned to the closest depot, from which it will then receive its supplies. To minimize shipping costs, the total distance sum, defined as

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must be as small as possible.

Write a program that computes the positions of the k depots, such that the total distance sum is minimized.

Input

The input file contains several descriptions of fastfood chains. Each description starts with a line containing the two integers n and k. n and k will satisfy 1 <= n <= 200, 1 <= k <= 30, k <= n. Following this will n lines containing one integer each, giving the positions di of the restaurants, ordered increasingly.

The input file will end with a case starting with n = k = 0. This case should not be processed.

Output

For each chain, first output the number of the chain. Then output a line containing the total distance sum.

Output a blank line after each test case.

Sample Input

6 3 5 6 12 19 20 27 0 0

Sample Output

Chain 1 Total distance sum = 8

一来没思路，看了题解，目前动态规划的题几乎都看了题解才有思路。。。。。。

思路：[i,j]之间加一个仓库，为了使要增加的路程最短，只能加在第(int)(i+j)/2个餐厅。

case[i][j]为在[i,j]之间加一个仓库所要增加的路程。

dp[i][j]表示在前j个餐厅之中加i个仓库所要走的路程。

状态转移方程：dp[i][j]=min(dp[i][j],dp[i-1][m]+case[m+1][j]); （i-1<=m<=j-1）

#include<iostream>
#include<cstdio>
#include<stdlib.h>
#include<cstring>
using namespace std;
#define LL long long

int dis[205];
int dist[205][205];
int dp[35][205];

int main()
{
    int n,k,cases=1;
    while(scanf("%d%d",&n,&k)!=EOF&&n&&k)
    {
        memset(dist,0,sizeof(dist));

        for(int i=0; i<=k; i++)
            for(int j=0; j<=n; j++)
                dp[i][j]=2000000000;
                //cout<<dp[0][0];
        dp[0][0]=0;
        dp[1][1]=0;
        for(int i=1; i<=n; i++)
            scanf("%d",&dis[i]);
        for(int i=1; i<=n-1; i++)
            for(int j=i; j<=n; j++)
            {
                int mid=(i+j)/2;
                for(int k=i; k<=j; k++)
                    dist[i][j]+=abs(dis[mid]-dis[k]);
            }
        for(int i=1;i<=n;i++)
            dp[1][i]=dist[1][i];
        for(int j=2; j<=n; j++)
            for(int i=2; i<=j,i<=k; i++)
                for(int m=i-1; m<=j-1; m++)
                    dp[i][j]=min(dp[i][j],dp[i-1][m]+dist[m+1][j]);

        printf("Chain %d\n",cases++);
        printf("Total distance sum = %d\n\n",dp[k][n]);
    }
    return 0;
}

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HDU_1227_Fast Food_动态规划

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原文地址：http://www.cnblogs.com/jasonlixuetao/p/5468589.html

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