UVA 662 Fast Food

Original question:
The Fastfood chain Mcburger owns several restaurants along a highway. Recently, they has decided to build several depots along the highway, each one located at a restaurent and supplying Seve Ral of the restaurants with the needed ingredients. Naturally, these depots should be placed so, the average distance between a restaurant and its assigned depot is Minim Ized. You is to write a program that computes the optimal positions and assignments of the depots. To do this more precise, the management of Mcburger have issued the following specification:you would be given the Positi ONS of n restaurants along the highway as n integers d 1 < D 2 < < D N (these is the distances measured from the company's headquarter, which happens to BES at the same
Highway). Furthermore, a number K (k≤n) would be given, and the number of depots to be built. The k depots is built at the locations of K different restaurants. Each restaurant'll is assigned to the closest depot and from which it'll then receive its supplies. To minimize shipping costs, the total distance sum, defined as N∑i=1 | D i− (position of depot serving restaurant i) | Must be as small as possible. Write A program this computes the positions of the K depots, such the total distance sum is minimized.
Input
The input file contains several descriptions of fastfood chains. Each description starts with a line containing the integers n and K. N and K would satisfy 1≤n≤200, 1≤k≤30, k≤ N. Following this would n lines containing one integer each, giving the positions D I of the restaurants, ordered Increasi ngly. The input file would end with a case starting with n = k = 0. This case is should not being processed.
Output
For each chain, first output the number of the chain. Then output an optimal placement of the depots as Follows:for, depot output a line containing its position and the RA Nge of Restaurants it serves. If there is more than one optimal solution, and output any of them. After the depot descriptions output a line containing the total distance sum, as defined in the problem text. Output a blank line after each test case.
Sample Input
6 3
5
6
12
19
20
27
0 0
Sample Output
Chain 1
Depot 1 at Restaurant 2 serves restaurants 1 to 3
Depot 2 at Restaurant 4 serves restaurants 4 to 5
Depot 3 at Restaurant 6 serves restaurant 6
Total Distance sum = 8

English:
give you two numbers n and K, respectively, that there are n stores, to be in the K store inventory. And then give you the number of n, indicating the coordinates of the n stores on a one-dimensional axis. Now ask you in this n store choose which K shop when warehouse can make this K store to the surrounding shops to transport goods in the shortest distance. Distance is expressed in absolute value.

#include <bits/stdc++.h> using namespace std;
int mark[201][201];
int dp[50][210];
int pos[210];
int x[201][201],y[201][201];

int n,k;
    int Get_mid_dis (int a[],int s,int e) {if (s==e) return 0;
    int mid= (e+s+1)/2;
    int tot=0;
    for (int i=s;i<=e;i++) tot+=abs (A[mid]-a[i]);
return tot;
    } void Ini () {memset (dp,0,sizeof (DP));
    memset (Mark,0,sizeof (Mark));
    memset (x,0,sizeof (x));
    memset (y,0,sizeof (y));
    for (int. i=1;i<=n;i++) {for (int j=i+1;j<=n;j++) Mark[i][j]=get_mid_dis (POS,I,J);
    }} int main () {Ios::sync_with_stdio (false);
    int t=1;

        while (Cin>>n>>k,n+k) {for (int i=1;i<=n;i++) cin>>pos[i];
        INI ();

        for (int i=2;i<=n;i++) dp[1][i]=mark[1][i];
                for (int j=2;j<=k;j++) {for (int i=j;i<=n;i++) {int tmp=int_max;
for (int t=j-1;t<i;t++)                {if (dp[j-1][t]+mark[t+1][i]<tmp) {tmp
                        =dp[j-1][t]+mark[t+1][i];
                        x[j][i]=t+1;
                    Y[j][i]=i;
            }} dp[j][i]=tmp;
        }} int i=n,j=k;
        Vector<int> Rx,ry;
                while (true) {if (j==1) {rx.push_back (j);
                Ry.push_back (i);
            Break
            } rx.push_back (X[j][i]);
            Ry.push_back (Y[j][i]);
            I=x[j][i]-1;
        j--;
        } cout<< "Chain" <<T++<<endl;
        I=1; for (auto Ritx=rx.rbegin (), Rity=ry.rbegin (); Ritx!=rx.rend (); ritx++,rity++,i++) if (*ritx!=*rity) cout&lt ;< "Depot" <<i<< "at Restaurant" << (*ritx+*rity)/2<< "serves restaurants" <<*ritx< < "to" <<*rity<&Lt;endl; else cout<< "Depot" <<i<< "at Restaurant" << (*ritx+*rity)/2<< "serves restaurant
        "<<*ritx<<endl;
    cout<< "Total distance sum =" <<dp[k][n]<<endl<<endl;
} return 0;
 }

Ideas:

A very good dynamic planning problem, better think. The state transition equation is
Dp[j][i]=min (Dp[j-1][k]+dis (K+1,J))

wherein Dis (I,J) represents the first store to the J store to set a warehouse supply distance, warehouse location is the median, and then accumulate I to J store coordinates to the sum of the median.
Dp[j][i] Indicates the minimum distance to set J warehouses in the first I store
Then the significance of the transfer equation is the first I store set J warehouse of the distance is equal to the former K store set J-1 Warehouse, and then add the new addition of the first J warehouse after the shortest distance.