• LeetCode 1066. Campus Bikes II


    原题链接在这里:https://leetcode.com/problems/campus-bikes-ii/

    题目:

    On a campus represented as a 2D grid, there are N workers and M bikes, with N <= M. Each worker and bike is a 2D coordinate on this grid.

    We assign one unique bike to each worker so that the sum of the Manhattan distances between each worker and their assigned bike is minimized.

    The Manhattan distance between two points p1 and p2 is Manhattan(p1, p2) = |p1.x - p2.x| + |p1.y - p2.y|.

    Return the minimum possible sum of Manhattan distances between each worker and their assigned bike.

    Example 1:

    Input: workers = [[0,0],[2,1]], bikes = [[1,2],[3,3]]
    Output: 6
    Explanation: 
    We assign bike 0 to worker 0, bike 1 to worker 1. The Manhattan distance of both assignments is 3, so the output is 6.
    

    Example 2:

    Input: workers = [[0,0],[1,1],[2,0]], bikes = [[1,0],[2,2],[2,1]]
    Output: 4
    Explanation: 
    We first assign bike 0 to worker 0, then assign bike 1 to worker 1 or worker 2, bike 2 to worker 2 or worker 1. Both assignments lead to sum of the Manhattan distances as 4.

    Note:

    1. 0 <= workers[i][0], workers[i][1], bikes[i][0], bikes[i][1] < 1000
    2. All worker and bike locations are distinct.
    3. 1 <= workers.length <= bikes.length <= 10

    题解:

    For each worker, check which is the cloest bike, assign it to the worker. And check the next worker until we hit the end of worker list.

    We could use state to record which bike has been used. say 01101 means bike 0, 2, 3 have been used because bit 0, 2, 3 are equal to 1.

    Check bike i, if state & (1 << i) != 0, this means the bike i has been taken before.

    Otherwise, we use this bike and accumlate the manhattan distance.

    DFS state needs workers, worker index, bikes, current state and dp to record state. DFS returns minimum distance sum with current state.

    Time Complexity: O(n!). n = bikes.length.

    Space: O(1 << n).

    AC Java:

     1 class Solution {
     2     public int assignBikes(int[][] workers, int[][] bikes) {
     3         int n = bikes.length;
     4         int [] dp = new int[1 << n];
     5         return dfs(workers, 0, bikes, 0, dp);
     6     }
     7     
     8     private int dfs(int [][] workers, int workerIndex, int [][] bikes, int state, int [] dp){
     9         if(workerIndex == workers.length){
    10             return 0;
    11         }
    12         
    13         if(dp[state] != 0){
    14             return dp[state];
    15         }
    16         
    17         int min = Integer.MAX_VALUE;
    18         for(int i = 0; i < bikes.length; i++){
    19             if((state & (1 << i)) == 0){
    20                 min = Math.min(min, dist(workers[workerIndex], bikes[i]) + dfs(workers, workerIndex + 1, bikes, state | (1 << i), dp));
    21             }
    22         }
    23         
    24         dp[state] = min;
    25         return min;
    26     }
    27     
    28     private int dist(int [] a, int [] b){
    29         return Math.abs(a[0] - b[0]) + Math.abs(a[1] - b[1]);
    30     }
    31 }

    类似Campus Bikes.

  • 相关阅读:
    余佳文 超级课程表创始人
    JavaScript 运行机制详解:再谈Event Loop
    koa2 async和await 实战详解
    node.js
    Mac版MySQL安装最新教程
    nw.js 打包与发布
    【514】keras Dense 层操作三维数据
    【513】keras 后端函数说明
    【512】keras 中的核心网络层
    【511】Keras 函数式 API
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/12204080.html
Copyright © 2020-2023  润新知