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UVA 662 Fast Food(DP)
2015-11-21 01:01:19 来源: 作者: 【 】 浏览:1
Tags:UVA 662 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 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.

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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 $k (k \le? n)$ 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

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

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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.

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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 $1 \le? n\le? 200$, $1 \le? k ?\le 30$, $k \le n$. 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.

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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.

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Output a blank line after each test case.

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Sample Input

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6 3
5
6
12
19
20
27
0 0

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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

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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 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
   
    
#include
    
      #include
     
       #include
      
        #include
       
         #include
        
          #include
         
           #include
          
            #include
            #include
            
              #include
             
               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
              
               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
               
              
             
            
          
         
        
       
      
     
    
   
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