题目

Description

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.

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 <= n <= 200, 1 <= k <= 30, k <= n. Following this will n lines containing one integer each, giving the positions di 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 a line containing the total distance sum.

Output a blank line after each test case.

Sample Input

6 3
5
6
12
19
20
27
0 0

Sample Output

Chain 1
Total distance sum = 8

题解

看上去毫无思路,不过按照动态规划的套路,应该是这样的解法
用 dp[i][j] 表示 i 个仓库,前 j 个饭店的最小值

那么新建的仓库应该建在 1 和 j 中间(选择 k 把原仓库控制区间分成两个)
如果用 Min[i][j] 表示在 i 和 j 中间建仓库的最小距离
应该有 dp[i][j] = min{ dp[i-1][k] + Min[k+1][j] }

如果是偶数个饭点,仓库应该建在中间两个饭店中间的任意位置
如果是奇数个饭店,仓库应该建在中间饭店的位置
这样可以保证距离最短

先用 O(n²) 的时间算出来 Min[i][j]
然后 O(n³) 的时间算出来 dp[i][j]

对于状态转移方程的正确性的解释
对于需要计算的 dp[i][j] 其必然是有 i-1 个仓库再加上一个仓库得到
可能存在的极端情况是,两端饭店离得非常远,为了保证最近,因此 i-1 将仓库建在中间位置
而可以建 i 个饭店时,可以将饭店分别建在两端
当 k 为最右面的左半部分饭店时,恰好将中间距离给跳掉,因此为最优解
对于多个仓库的情况,可以将其类比为该种情况

代码

/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <functional>
#include <bitset>
using namespace std;

const int INF = 0x7FFFFFFF / 2;

const int maxn = 205;
const int maxk = 35;

int kase = 1;

int restaurants[maxn];
int dp[maxk][maxn];
int Min[maxn][maxn];

bool Do() {
    memset(dp,0,sizeof(dp));

    int n,m;
    cin >> n >> m;

    if(n == 0 && m == 0)
        return false;

    for(int i = 1;i <= n;i++)
        cin >> restaurants[i];

    for(int i = 1;i <= n;i++) {
        for(int j = i;j <= n;j++) {
            if(i == j) {
                Min[i][j] = 0;
            } else {
                int pos = restaurants[(i + j) / 2];
                Min[i][j] = 0;
                for(int k = i;k <= j;k++) {
                    Min[i][j] += abs(pos - restaurants[k]);
                }
            }
        }
    }

    for(int i = 1;i <= m;i++)
        if(i == 1) {
            for(int j = 1;j <= n;j++)
                dp[1][j] = Min[1][j];
        } else {
            for(int j = 1;j <= n;j++) {
                dp[i][j] = INF;
                for(int k = 1;k < j;k++) {
                    dp[i][j] = min(dp[i][j],dp[i - 1][k] + Min[k + 1][j]);
                }
            }
        }
    cout << "Chain "<<kase++ << endl <<"Total distance sum = "<<dp[m][n] << endl<<endl;

    return true;
}
int vs_main() {
    cin.tie(0);
    std::cin.sync_with_stdio(false);

    while(Do());
    return 0;
}