題意是在n個餐廳之間選擇k個地方修建倉庫使得，各個餐廳到倉庫的距離和最小。

dp[i][j] = min{dp[i][j], dp[i-1][m]+w[m+1][j]}。i表示倉庫數目，j表示第幾個餐廳。狀態轉移方程式的意思是：到達j的時候區間(i,j)上所擁有的最小的值。到達j的時候可以選擇j位置上不修建倉庫dp[i][j] = dp[i][j]。或者選擇修倉庫那麼之前已經有i-1個倉庫修建好了，所以只能從i-1到j-1之間選擇修建第i個倉庫，所以從i-1->j-1列舉m。然後加上一個增量w[m+1][j]。表示加上i後這一段上增加的值的最小。

Fast Food

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. 

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-7
#define M 10001000
//#define LL __int64
#define LL long long
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
const int maxn = 101000;

using namespace std;

int dp[500][500];
int w[500][500];
int num[500];

int main()
{
    int k, n;
    int Case = 1;
    while(cin >>n>>k)
    {
        if(!n && !k)
            break;
        memset(dp, INF, sizeof(dp));
        memset(w, 0 , sizeof(w));
        memset(num, 0, sizeof(num));
        for(int i = 1; i <= n; i++)
            cin >>num[i];
        for(int i = 1; i <= n; i++)
            for(int j = i; j <= n; j++)
                for(int m = i; m <= j; m++)
                    w[i][j] += abs(num[m]-num[(i+j)/2]);
        for(int i = 1; i <= n; i++)
            dp[1][i] = w[1][i];
        for(int i = 2; i <= k; i++)
            for(int j = i; j <= n; j++)
                for(int m = i-1; m <= j-1; m++)
                    dp[i][j] = min(dp[i][j], dp[i-1][m]+w[m+1][j]);
        cout<<"Chain "<<Case++<<endl;
        cout<<"Total distance sum = "<<dp[k][n]<<endl<<endl;
    }
    return 0;
}