128th LeetCode Weekly Contest Capacity To Ship Packages Within D Days

最新推荐文章于 2025-04-16 09:45:00 发布

转载最新推荐文章于 2025-04-16 09:45:00 发布 · 121 阅读

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原文链接：http://www.cnblogs.com/yinghualuowu/p/10549721.html

本文探讨了一种算法问题，即确定最小的船载重量，使所有货物能在指定天数内从一个港口运送到另一个港口。通过使用二分查找法，文章提供了一个有效的解决方案，包括获取最大值、总和及所需船只数量的函数。

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 <= D <= weights.length <= 50000
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);
    }
};

转载于:https://www.cnblogs.com/yinghualuowu/p/10549721.html