邮局选址问题

链接：https://www.nowcoder.com/questionTerminal/8b0152848ffd475eaa950278606fe70b
来源：牛客网
一条直线上有居民点，邮局只能建在居民点上。给定一个有序整形数组arr，每个值表示居民点的一维坐标，再给定一个正数num，表示邮局数量。 选择num个居民点建立num个邮局，使所有的居民点到邮局的总距离最短，返回最短的总距离。

import java.util.Scanner;

public class Main {

    private static int[][] getRecord(int[] arr) {
        int[][] record = new int[arr.length][arr.length];
        for (int i = 0; i < arr.length; ++i) {
            for (int j = i + 1; j < arr.length; ++j) {
                record[i][j] = record[i][j - 1] + arr[j] - arr[(i + j) >> 1];
            }
        }
        return record;
    }

    /**
     * 未优化版本
     *
     * @param arr
     * @param m
     * @return
     */
    private static int solve(int[] arr, int m) {
        if (arr == null || arr.length == 0) {
            return 0;
        }

        int n = arr.length;
        int[][] record = getRecord(arr);
        int[][] dp = new int[n][n];

        for (int i = 0; i < n; ++i) {
            dp[i][0] = record[0][i];
        }

        for (int i = 1; i < n; ++i) {
            for (int j = 1; j < Math.min(m, i + 1); ++j) {
                dp[i][j] = record[0][i];
                for (int s = i; s > 0; --s) {
                    dp[i][j] = Math.min(dp[i][j], dp[s - 1][j - 1] + record[s][i]);
                }
            }
        }

        return dp[n - 1][m - 1];
    }

    /**
     * 四边形不等式优化
     *
     * @param arr
     * @param m
     * @return
     */
    private static int solvePlus(int[] arr, int m) {
        if (arr == null || arr.length == 0) {
            return 0;
        }

        int n = arr.length;
        int[][] record = getRecord(arr);
        int[][] dp = new int[n][n];
        int[][] choose = new int[n][n];

        for (int i = 0; i < n; ++i) {
            dp[i][0] = record[0][i];
        }

        for (int i = 1; i < n; ++i) {
            for (int j = Math.min(m - 1, i); j >= 1; --j) {
                dp[i][j] = record[0][i];
                int up = j == Math.min(m - 1, i) ? i : Math.min(i, choose[i][j + 1]);
                int down = Math.max(1, choose[i - 1][j]);

                for (int s = up; s >= down; --s) {
                    if (dp[i][j] > dp[s - 1][j - 1] + record[s][i]) {
                        dp[i][j] = dp[s - 1][j - 1] + record[s][i];
                        choose[i][j] = s;
                    }
                }
            }
        }

        return dp[n - 1][m - 1];
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        while (in.hasNext()) {
            int n = in.nextInt();
            int m = in.nextInt();
            int[] arr = new int[n];
            for (int i = 0; i < n; ++i) {
                arr[i] = in.nextInt();
            }
            System.out.println(solvePlus(arr, m));
        }
    }
}

标签：arr,record,邮局,++,问题,int,length,选址,dp
来源： https://www.cnblogs.com/tianyiya/p/15428826.html