Minimum Cost for Cutting Cake II

Description

There is an m x n cake that needs to be cut into 1 x 1 pieces.

You are given integers m, n, and two arrays:

• horizontalCut of size m - 1, where horizontalCut[i] represents the cost to cut along the horizontal line i.
• verticalCut of size n - 1, where verticalCut[j] represents the cost to cut along the vertical line j.

In one operation, you can choose any piece of cake that is not yet a 1 x 1 square and perform one of the following cuts:

1. Cut along a horizontal line i at a cost of horizontalCut[i].
2. Cut along a vertical line j at a cost of verticalCut[j].

After the cut, the piece of cake is divided into two distinct pieces.

The cost of a cut depends only on the initial cost of the line and does not change.

Return the minimum total cost to cut the entire cake into 1 x 1 pieces.

Example 1:

Input: m = 3, n = 2, horizontalCut = [1,3], verticalCut = [5]

Output: 13

Explanation:

• Perform a cut on the vertical line 0 with cost 5, current total cost is 5.
• Perform a cut on the horizontal line 0 on 3 x 1 subgrid with cost 1.
• Perform a cut on the horizontal line 0 on 3 x 1 subgrid with cost 1.
• Perform a cut on the horizontal line 1 on 2 x 1 subgrid with cost 3.
• Perform a cut on the horizontal line 1 on 2 x 1 subgrid with cost 3.

The total cost is 5 + 1 + 1 + 3 + 3 = 13.

Example 2:

Input: m = 2, n = 2, horizontalCut = [7], verticalCut = [4]

Output: 15

Explanation:

• Perform a cut on the horizontal line 0 with cost 7.
• Perform a cut on the vertical line 0 on 1 x 2 subgrid with cost 4.
• Perform a cut on the vertical line 0 on 1 x 2 subgrid with cost 4.

The total cost is 7 + 4 + 4 = 15.

Constraints:

• 1 <= m, n <= 105
• horizontalCut.length == m - 1
• verticalCut.length == n - 1
• 1 <= horizontalCut[i], verticalCut[i] <= 103

Code

Time Complexity: $O(n\log n)$, Space Complexity: $O(1)$

Minimum Cost for Cutting Cake I 的 dp 是 4 維，所以在這裡會 TLE。

greedy：最後一定都會砍到，但是越切，方塊會越多，因此要先切 cost 大的。

class Solution {
public:
    long long minimumCost(int m, int n, vector<int>& horizontalCut, vector<int>& verticalCut) {
        using ll = long long;
        sort(horizontalCut.begin(), horizontalCut.end());
        sort(verticalCut.begin(), verticalCut.end());
 
        int hCount = 1, vCount = 1;
        ll res = 0;
        while(!horizontalCut.empty() || !verticalCut.empty()) {
            ll hcut = -1, vcut = -1;
            if(!horizontalCut.empty()) {
                hcut = horizontalCut.back();
            }
 
            if(!verticalCut.empty()) {
                vcut = verticalCut.back();
            }
 
            if(vcut >= hcut) {
                res += vcut * hCount;
                verticalCut.pop_back();
                vCount++;
            } else {
                res += hcut * vCount;
                hCount++;
                horizontalCut.pop_back();
            }
            
        }
        return res;
    }
};

Source

• Minimum Cost for Cutting Cake II - LeetCode