hdu 1227 Fast Food(DP线性决策)

Fast Food

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1403    Accepted Submission(s): 609

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

 

 

Recommend

JGShining

 

思路：f[a][b] a到b之间只有一个补给站时的最优路径长。dp[a][b]前b个店设有a个补给站的最优值。

f[a][b]=min(f[a-1][c]+f[c+1][b])(a-1<=c<=b-1)

 

#include<iostream>
#include<cstring>
using namespace std;
const int mm=210;
const int oo=1e9;
int f[mm][mm],dp[33][mm],data[mm];
int m,n;
inline int fabs(int x)
{
    return x>0?x:-x;
}
int main()
{   int cas=0;
    while(cin>>m>>n)
    {   if(m==0&&n==0)break;
        for(int i=1;i<=m;i++)cin>>data[i];
        memset(f,0,sizeof(f));
        for(int i=1;i<=m;i++)
        for(int j=i+1;j<=m;j++)
        {   for(int k=i;k<=j;k++)
            f[i][j]+=fabs(data[(i+j)/2]-data[k]);///i到j中只设一个补给站的最优值
        }
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        for(int j=i;j<=m;j++)
        dp[i][j]=oo;
        for(int i=1;i<=m;i++)///边界
        dp[1][i]=f[1][i];
        for(int i=2;i<=n;i++)///补给站的个数
        for(int j=i;j<=m;j++)///店的个数
        { for(int k=i-1;k<j;k++)
            {int z=dp[i-1][k]+f[k+1][j];
            ///cout<<z<<"z"<<i<<" "<<j<<endl;
             dp[i][j]=z<dp[i][j]?z:dp[i][j];
            }
        }
        cout<<"Chain "<<++cas<<"\n";
        cout<<"Total distance sum = "<<dp[n][m]<<"\n";
        cout<<"\n";
    }
}