Algorithm: Backtracking (回溯) 通常用于暴力列举、排列组合灯，注意 Pruning以优化 （LC 1240）

LC 1240 Tiling a Rectangle with the Fewest Squares

Given a rectangle of size n x m, find the minimum number of integer-sided squares that tile the rectangle.

Example 1:

Input: n = 2, m = 3
Output: 3
Explanation: 3 squares are necessary to cover the rectangle.
2 (squares of 1x1)
1 (square of 2x2)

Example 2:

Input: n = 5, m = 8
Output: 5

Example 3:

Input: n = 11, m = 13
Output: 6

 

Constraints:

• 1 <= n <= 13
• 1 <= m <= 13

 

思考过程：一开始思考，往trick方向想，觉得要不要有特殊的算法，先取最小边，然后填上最小边长的正方形等等。后来发现，想复杂了。这道题只需要暴力枚举，注意剪枝和优化即可。

 

 

class Solution {
public: 
    int tilingRectangle(int n, int m) {
        int size = max(n, m);
        vector height(size, 0);
        int total = 0;
        dfs(total, height, min(n, m));
        return global_total_min_;
    }
    
private:
    void dfs(int total, vector height, const int target_height) {
        if (total > global_total_min_) {
            return;
        }

        int smallest_index = -1;
        
        for (int i = 0; i < height.size(); ++i) {
            if (height.at(i) == target_height) {
                continue;
            }
            if (smallest_index != -1) {
                smallest_index = height.at(smallest_index) > height.at(i) ? i : smallest_index;
            } else {
                smallest_index = i;
            }
        }
        
        if (smallest_index != -1) {
            // continue to fullfill
            int local_start = smallest_index;
            int local_end = smallest_index + 1;
            
            while (local_end != height.size() && 
                   height.at(local_end) == height.at(local_start)) {
                ++local_end;
            }
            
            // 错误1：忘了回溯需要回复原样
            vector backup_height = height;
            
            for (; local_end != local_start; --local_end) {
                int next_square_size = min(local_end - local_start, 
                                       target_height - height.at(smallest_index));
            
                height = backup_height;
                // 错误2： 忘了这个方形是等长，终止条件没设对
                for (int i = local_start; i < local_start + next_square_size; ++i) {
                    // 错误3：忘了是叠加，不是重设；
                    height[i] += next_square_size;
                }

                long pattern = height.at(0);
                // calculate the current pattern
                for (int i = 1; i < height.size(); ++i) {
                    pattern = (pattern * target_height) + height.at(i);
                }

                if (pattern_met_min_.find(pattern) != pattern_met_min_.end() && 
                    pattern_met_min_.at(pattern) < total) {
                    // 错误4，不是遇到重复的pattern就返回，还要对比局部最优
                    // met before, don't need to do again
                    return;
                }
                
                pattern_met_min_[pattern] = total;
                
                dfs(total + 1, height, target_height);
            }
        } else {
          // a full status
          global_total_min_ = min(total, global_total_min_);
        }
    }
    
    unordered_map pattern_met_min_;
    int global_total_min_ = INT_MAX;
};