UVA 662 Fast Food(DP)

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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 restaurent 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 $d_1 < d_2 < \dots < d_n$ (these are the distances measured from the company's headquarter, which happens to be at the same highway). Furthermore, a number $k (k \leŸ n)$ 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

$\begin{displaymath}\sum_{i=1}^n \mid d_i - (\mbox{position of depot serving restaurant }i) \mid\end{displaymath}$

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 $1 \leŸ n\leŸ 200$ , $1 \leŸ k Ÿ\le 30$ , $k \le n$ . Following this will n lines containing one integer each, giving the positions d_i 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 an optimal placement of the depots as follows: for each depot output a line containing its position and the range of restaurants it serves. If there is more than one optimal solution, 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 356121920270 0

Sample Output

Chain 1Depot 1 at restaurant 2 serves restaurants 1 to 3Depot 2 at restaurant 4 serves restaurants 4 to 5Depot 3 at restaurant 6 serves restaurant 6Total distance sum = 8

考虑dp状态为dp[i][j]:前i个酒店建k个仓库的最小代价。dp[i][j]=min(dp[k][j-1]+dis[k+1][i])(j-1<=k<i);关于路径输出设path[i][j]为达到dp[i][j]状态的前一状态的末仓库。易知道path[i][1]=0;其他的就是递归输出了。

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<iostream>#include<queue>#include<cmath>#include<map>#include<stack>#include<bitset>using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )typedef long long LL;typedef pair<int,int>pil;const int INF = 0x3f3f3f3f;const int maxn=220;int dis[maxn][maxn];int dp[maxn][50];int path[220][50];int d[220];int n,k,cnt;int cal(int l,int r){    int sum=0;    int mid=(l+r)>>1;    for(int i=l;i<=r;i++)        sum+=abs(d[mid]-d[i]);    return sum;}void print(int i,int x){    if(x==0)        return ;    print(path[i][x],x-1);    printf("Depot %d at restaurant %d serves restaurants %d to %d\n",x,(path[i][x]+1+i)/2,path[i][x]+1,i);}int main(){    int cas=1;    while(~scanf("%d%d",&n,&k)&&(n+k))    {        CLEAR(dp,INF);        CLEAR(path,0);cnt=1;        REPF(i,1,n)  scanf("%d",&d[i]);        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)               dis[i][j]=cal(i,j);        for(int i=1;i<=n;i++)        {            dp[i][1]=dis[1][i];            path[i][1]=0;        }        for(int i=1;i<=n;i++)        {            for(int j=2;j<=k;j++)            {                for(int p=j-1;p<i;p++)                    if(dp[p][j-1]+dis[p+1][i]<dp[i][j])                    {                        dp[i][j]=dp[p][j-1]+dis[p+1][i];                        path[i][j]=p;//上个区间                    }            }        }        printf("Chain %d\n",cas++);        print(n,k);        printf("Total distance sum = %d\n\n",dp[n][k]);    }    return 0;}/**/

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