zoukankan      html  css  js  c++  java
  • 128th LeetCode Weekly Contest Capacity To Ship Packages Within D Days

    A conveyor belt has packages that must be shipped from one port to another within D days.

    The i-th package on the conveyor belt has a weight of weights[i].  Each day, we load the ship with packages on the conveyor belt (in the order given by weights). We may not load more weight than the maximum weight capacity of the ship.

    Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within D days.

    Example 1:

    Input: weights = [1,2,3,4,5,6,7,8,9,10], D = 5
    Output: 15
    Explanation: 
    A ship capacity of 15 is the minimum to ship all the packages in 5 days like this:
    1st day: 1, 2, 3, 4, 5
    2nd day: 6, 7
    3rd day: 8
    4th day: 9
    5th day: 10
    
    Note that the cargo must be shipped in the order given, so using a ship of capacity 14 and splitting the packages into parts like (2, 3, 4, 5), (1, 6, 7), (8), (9), (10) is not allowed. 
    

    Example 2:

    Input: weights = [3,2,2,4,1,4], D = 3
    Output: 6
    Explanation: 
    A ship capacity of 6 is the minimum to ship all the packages in 3 days like this:
    1st day: 3, 2
    2nd day: 2, 4
    3rd day: 1, 4
    

    Example 3:

    Input: weights = [1,2,3,1,1], D = 4
    Output: 3
    Explanation: 
    1st day: 1
    2nd day: 2
    3rd day: 3
    4th day: 1, 1
    

    Note:

    1. 1 <= D <= weights.length <= 50000
    2. 1 <= weights[i] <= 500

    题意看错了,应该是找最大值。这样妥妥的二分啊

    当sum>max时候,我们开始二分。

    疯狂找(ai+....+ax) > mid 这个有几段 getRequiredPainters这个就这么用的

    int getMax(int A[], int n) {
        int max = INT_MIN;
        for (int i = 0; i < n; i++) {
            if (A[i] > max) max = A[i];
        }
        return max;
    }
    
    int getSum(int A[], int n) {
        int total = 0;
        for (int i = 0; i < n; i++)
            total += A[i];
        return total;
    }
    
    int getRequiredPainters(int A[], int n, int maxLengthPerPainter) {
        int total = 0, numPainters = 1;
        for (int i = 0; i < n; i++) {
            total += A[i];
            if (total > maxLengthPerPainter) {
                total = A[i];
                numPainters++;
            }
        }
        return numPainters;
    }
    
    
    
    int BinarySearch(int A[], int n, int k) {
        int lo = getMax(A, n);
        int hi = getSum(A, n);
    
        while (lo < hi) {
            int mid = lo + (hi-lo)/2;
            int requiredPainters = getRequiredPainters(A, n, mid);
            if (requiredPainters <= k)
                hi = mid;
            else
                lo = mid+1;
        }
        return lo;
    }
    class Solution {
    public:
        int shipWithinDays(vector<int>& weights, int D) {
            int num[50100] = {0};
            int cnt = 0;
            int len = weights.size();
            for(int i = 0 ; i < len ; i++){
                num[i] = weights[i]; 
            }
            return BinarySearch(num ,len, D);
        }
    };
  • 相关阅读:
    Unix系统编程(四)creat系统调用
    Unix系统编程(三)通用的I/O
    Unix系统编程(二)open的练习
    FTP协议的粗浅学习--利用wireshark抓包分析相关tcp连接
    Linux上的ftp服务器 vsftpd 之配置满天飞--设置匿名用户访问(不弹出用户名密码框)以及其他用户可正常上传
    intelj idea编译项目报错,Error:ajc: The method getDestHost() is undefined
    oracle索引优化
    wireshark 表达式备忘录
    rabbitmq日志记录进出的每条消息
    powerDesigner根据sql脚本来逆向生成pdm等模型
  • 原文地址:https://www.cnblogs.com/yinghualuowu/p/10549721.html
Copyright © 2011-2022 走看看