LC 807. Max Increase to Keep City Skyline-优快云博客

本文探讨了一个算法问题：如何在不改变二维网格城市从四个方向观察的天际线情况下，最大化增加建筑物的高度总和。通过保存横竖最大值的最小值并减去当前值的方法，实现了这一目标。示例中输入的网格经过调整后，天际线保持不变，而建筑物高度的总增量达到35。

In a 2 dimensional array grid, each value grid[i][j] represents the height of a building located there. We are allowed to increase the height of any number of buildings, by any amount (the amounts can be different for different buildings). Height 0 is considered to be a building as well.

At the end, the "skyline" when viewed from all four directions of the grid, i.e. top, bottom, left, and right, must be the same as the skyline of the original grid. A city's skyline is the outer contour of the rectangles formed by all the buildings when viewed from a distance. See the following example.

What is the maximum total sum that the height of the buildings can be increased?

Example:
Input: grid = [[3,0,8,4],[2,4,5,7],[9,2,6,3],[0,3,1,0]]
Output: 35
Explanation: 
The grid is:
[ [3, 0, 8, 4], 
  [2, 4, 5, 7],
  [9, 2, 6, 3],
  [0, 3, 1, 0] ]

The skyline viewed from top or bottom is: [9, 4, 8, 7]
The skyline viewed from left or right is: [8, 7, 9, 3]

The grid after increasing the height of buildings without affecting skylines is:

gridNew = [ [8, 4, 8, 7],
            [7, 4, 7, 7],
            [9, 4, 8, 7],
            [3, 3, 3, 3] ]

Notes:

1 < grid.length = grid[0].length <= 50.
All heights grid[i][j] are in the range [0, 100].
All buildings in grid[i][j] occupy the entire grid cell: that is, they are a 1 x 1 x grid[i][j]rectangular prism.

Runtime: 8 ms, faster than 49.01% of C++ online submissions for Max Increase to Keep City Skyline.

保存横竖最大值的最小值，减去当前值即可。

#include <algorithm>
class Solution {
public:
  int maxIncreaseKeepingSkyline(vector<vector<int>>& grid) {
    int r = grid.size();
    int c = grid[0].size();
    vector<int> rgrid(r,0);
    vector<int> cgrid(c,0);
    for(int i=0; i<r; i++) rgrid[i] = *max_element(grid[i].begin(), grid[i].end());
    for(int i=0; i<c; i++){
      int maxval = 0;
      for(int j=0; j<r; j++){
        maxval = max(maxval, grid[j][i]);
      }
      cgrid[i] = maxval;
    }
    //for(auto x : cgrid) cout << x << endl;
    //for(auto x : rgrid) cout << x << endl;
    int ret = 0, tmp = 0;
    for(int i=0; i<r; i++){
      for(int j=0; j<c; j++){
        tmp = min(rgrid[i],cgrid[j]) - grid[i][j];
        //cout << tmp << " " << grid[i][j] << endl;
        ret += tmp > 0 ? tmp : 0;
      }
    }
    return ret;
  }
};