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:
    [No0000C9]神秘的掐指一算是什么?教教你也会
    [No0000C8]英特尔快速存储IRST要不要装
    [No0000C7]windows 10桌面切换快捷键,win10
    [No0000C6]Visual Studio 2017 函数头显示引用个数
    [No0000C4]TortoiseSVN配置外部对比工具
    [No0000C5]VS2010删除空行
    [No0000C3]StarUML2 全平台破解方法
    [No0000C2]WPF 数据绑定的调试
    [No0000C1]Excel 删除空白行和空白列VBA代码
    [No0000C0]百度网盘真实地址解析(不用下载百度网盘)20170301
  • 原文地址:https://www.cnblogs.com/yinghualuowu/p/10549721.html
Copyright © 2011-2022 走看看