Uva--662（动规，经典递推）

2014-08-10 23:30:04

  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.

To make this more precise, the management of McBurger has issued the following specification: You will be given the positions of nrestaurants along the highway as n integers  (these are the distances measured from the company's headquarter, which happens to be at the same highway). Furthermore, a number  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 kwill satisfy , , . 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 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

思路：这道题和之前的递推题是有异曲同工之妙的（类似的题目貌似在某个比赛里见过）。我一开始的思路是dp[i][j][k]记录从第i个到第j个饭店布置k个仓库的最小距离和，后来发现状态太多不好转移。
于是把表示区间的两个量压缩成一个，用dp[i][j]表示在前j个饭店布置i个仓库的最小距离和（这种前i个放j个东西的递推非常常见）。为什么要这样想呢？在分析一下：首先动态规划的核心思想就是通过
求出子问题在一定条件的下的最优解，来推出整个问题的最优解，特点是：子问题具有重复性（或者说子问题中包含着更子的问题，因为如果子问题相互独立就可用分治来处理），其次是子问题和整个问题可
建立递推关系，就像这题中的：在1 - j号饭店中放i - 1个仓库，与在（j+1） - n号饭店中放1个仓库，这两个子问题想加的最优解是可以构成整个问题的最优解的。
　　于是有：dp[i][j] = min(dp[i][j],dp[i - 1][p] + dis[p + 1][j]) (i - 1 <= p < j)
　　(这里的dis[x][y]表示在x - y号饭店中布置一个仓库的最小距离和，显然应该放在中间，若为饭店数为偶数，随便偏左偏右，易证）

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
using namespace std;
const int INF = 1e9;

int n,k;
int d[205];
int dp[35][205];
int dis[205][205];
int path[35][205];
int cnt;

int Cal(int x,int y){
    int sum = 0;
    int mid = (x + y) / 2;
    for(int i = x; i <= y; ++i)
        sum += abs(d[i] - d[mid]);
    return sum;
}

void Print(int x,int y){
    if(x == 0)
        return;
    Print(x - 1,path[x][y]);
    printf("Depot %d at restaurant %d serves restaurants %d to %d\n",++cnt,(path[x][y] + 1 + y) / 2,path[x][y] + 1,y);
}

int main(){
    int Case = 0;
    while(scanf("%d%d",&n,&k) == 2 && (n || k)){
        for(int i = 1; i <= n; ++i)
            scanf("%d",&d[i]);
        for(int i = 1; i <= n; ++i){
            for(int j = i; j <= n; ++j){
                dis[i][j] = Cal(i,j);
            }
        }
        //dp[i][j] = min(dp[i][j],dp[i - 1][p] + dis[p + 1][j]) , (i - 1 <= p < j)
        memset(dp,0,sizeof(dp));
        memset(path,0,sizeof(path));
        for(int j = 1; j <= n; ++j){
            dp[1][j] = dis[1][j];
        }
        for(int i = 2; i <= k; ++i){
            for(int j = 1; j <= n; ++j){
                dp[i][j] = INF;
                for(int p = i - 1; p < j; ++p){
                    if(dp[i - 1][p] + dis[p + 1][j] < dp[i][j]){
                        dp[i][j] = dp[i - 1][p] + dis[p + 1][j];
                        path[i][j] = p;
                    }
                }
            }
        }
        printf("Chain %d\n",++Case);
        cnt = 0;
        Print(k,n);
        printf("Total distance sum = %d\n\n",dp[k][n]);
    }
    return 0;
}