代码片段_Post Office Problem

There are n houses on a line. Given an array A and A[i] represents the position of i-th house. Now you need to pick k position to build k post offices.

What is the minimum sum distance from these n houses to the nearest post office?

All positions are integers.

Example

Example 1:

Input: A = [1, 2, 3, 4, 5], k = 2
Output: 3
Explanation: Build post offices on position 2 and 4.

Example 2:

Input: A = [1, 2], k = 1
Output: 1
Explanation: Build post office on position 1 or 2.

Challenge

O( $n^2??) time$

$思路：$

线性动态规划 (而不是区间动态规划)

可以按照每个房子最近的邮局, 把 n 个房子分成 k 段, 而我们要决定的就是这 k 段分别是多长. 为了处理方便我们先对房子的位置排序.

设定 f[i][j] 表示前 j 栋房子建立 i 个邮局时的最优解. 对于这个状态我们需要决策的就是 j 之前有多少栋房子共用第 i 个邮局, 故有:

f[i][j] = min{f[i - 1][j - x] + sumdis[j - x][j - 1]}
其中 sumdis[l][r] 表示下标范围为 [l, r] 的房子之间建立一个邮局, 这些房子与该邮局的最短距离
(注意f[i][j]中的j表示的第j栋房子从1计数, sumdis从0计数)

sumdis数组可以实现预处理出来, 具体算法与中位数的性质有关. 即对于 sumdis[l][r], 直接选择这 r - l + 1 栋房子中, 中间的那一栋建立邮局(如果是偶数栋, 中间的两栋任选一栋), 这时这些房子与邮局的距离之和是最短的.

至于dp的边界: f[i][0] = 0, f[0][j] = INF, 以及 i >= j 时 f[i][j] = 0

另外, 这样的状态定义可以用滚动数组优化空间.

public class Solution {
    /**
     * @param A an integer array
     * @param k an integer
     * @return an integer
     */
    int[][] init(int[] A) {
        int n = A.length;
        int[][] dis = new int[n + 1][n + 1];
        for (int i = 1; i <= n; i++) {
            for (int j = i + 1; j <= n; ++j) {
                int mid = (i + j) / 2;
                for (int k = i; k <= j; ++k)
                    dis[i][j] += Math.abs(A[k - 1] - A[mid - 1]);
            }
        }
        return dis;
    }

    public int postOffice(int[] A, int k) {
        // Write your code here
        int n = A.length;
        Arrays.sort(A);

        int[][] dis = init(A);
        int[][] dp = new int[n + 1][k + 1];
        if (n == 0 || k >= A.length)
            return 0;
        int ans = Integer.MAX_VALUE;
        for (int i = 0; i <= n; ++i) {
            dp[i][1] = dis[1][i];

        }

        for (int nk = 2; nk <= k; nk++) {
            for (int i = nk; i <= n; i++) {
                dp[i][nk] = Integer.MAX_VALUE;
                for (int j = 0; j < i; j++) {
                    if (dp[i][nk] == Integer.MAX_VALUE || dp[i][nk] > dp[j][nk - 1] + dis[j + 1][i])
                        dp[i][nk] = dp[j][nk - 1] + dis[j + 1][i];
                }
            }
        }
        return dp[n][k];
    }
}

Post Office Problem

Description

Example

Challenge