首页 > 代码库 > [ACM] HDU 1227 Fast Food （经典Dp）

[ACM] HDU 1227 Fast Food （经典Dp）

2024-07-26 05:33:32   223人阅读

Fast Food

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2173    Accepted Submission(s): 930


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


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
 


 

Source
Southwestern Europe 1998

 

解题思路：

题意为 一条路上有 n个商店，每个商店有一个x坐标位置，有k个仓库，要把k个仓库安放在n个位置中的k个上面，每个商店都向最近的仓库来获得补给，求怎么安放这k个仓库，使得每个商店到对应仓库的距离只和加起来最小，输出最小值。

解决本题要意识到两点:

1. 如果要在第i个位置和第j个位置之间安放仓库，那么要把它安放在  （i+j)/2 个位置上，（i+j)/2为整数, 才能保证从i到j个商店到仓库之间的距离之和最短。

2.如果按照题意把k个仓库安放在n个位置上,使得距离和最短，这样求得了最小值， 那么一定符合题意：每个商店都向最近的仓库来获得补给，因为如果不是向最近的，距离和肯定不是最短

用dp[i][j] 代表 前j个商店，有i个仓库 

那么状态转移方程为：

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

dp[i][j] 要从前一个状态推出来，及前k个商店有i-1个仓库，k是不确定的，但可以确定它的范围，最小是i-1 （一个商店位置上放一个仓库）,最大是j-1 ( 把第i个仓库放在第j个位置上） ,    cost [ i ] [ j ]为在i ,j之间放一个仓库的最小距离和，即前面提到的第1点。

 

代码：

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <cmath>
#include <string.h>
using namespace std;
int n,k;
int dp[32][210];//dp[i][j]前j个商店有i个仓库
int dis[210];
const int inf=0x3f3f3f3f;

int cost(int i,int j)
{
    int ans=0;
    int mid=dis[(i+j)/2];
    for(int k=i;k<=j;k++)
        ans+=abs(dis[k]-mid);
    return ans;
}

int main()
{
    int c=1;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        if(!n||!k)
            break;
        for(int i=1;i<=n;i++)
            scanf("%d",&dis[i]);
        memset(dp,inf,sizeof(dp));
        for(int i=1;i<=n;i++)
            dp[1][i]=cost(1,i);
        for(int i=2;i<=k;i++)//第i个仓库
            for(int j=1;j<=n;j++)//前j个商店
            {
                for(int k=i-1;k<=j-1;k++)
                    dp[i][j]=min(dp[i][j],dp[i-1][k]+cost(k+1,j));
            }

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


 

 

[ACM] HDU 1227 Fast Food （经典Dp）


联系
我们