【狀壓 dp】A000_LC_轉化爲全零矩陣的最少反轉次數（回溯 / dp / bfs）

一、Problem

Given a m x n binary matrix mat. In one step, you can choose one cell and flip it and all the four neighbours of it if they exist (Flip is changing 1 to 0 and 0 to 1). A pair of cells are called neighboors if they share one edge.

Return the minimum number of steps required to convert mat to a zero matrix or -1 if you cannot.

Binary matrix is a matrix with all cells equal to 0 or 1 only.

Zero matrix is a matrix with all cells equal to 0.

Input: mat = [[0,0],[0,1]]
Output: 3
Explanation: One possible solution is to flip (1, 0) then (0, 1) and finally (1, 1) as shown.

二、Solution

方法一：回溯

• 結束條件：
  • 當矩陣出現全 0 時，結算最小操作次數 min 並返回。
  • 當把所有位置都翻轉了一遍後，也返回。
• 本層遞歸的責任：對於當前格子，可選擇翻轉或不翻轉：
  • 選擇翻轉，那麼四周的格子也需要翻轉。
  • 否則，下一輪選擇。

這裏的換行技巧要熟練…

class Solution {
	int tot, R, C, min = Integer.MAX_VALUE, g[][];
	final static int[][] d = { {1,0},{0,-1},{0,1},{-1,0} };
	void dfs(int x, int y, int s) {
		boolean allZero = true;
		loop:
		for (int i = 0; i < R; i++)
		for (int j = 0; j < C; j++) {
			if (g[i][j] != 0) {
				allZero = false;
				break loop;
			}
		}
		if (allZero) {
			min = Math.min(min, s);
			return;
		}
		
		int nx = x, ny = y + 1;
		if (ny >= C) {
			nx++;
			ny = 0;
		}
        if (nx >= R)
            return;
		dfs(nx, ny, s);
		g[nx][ny] ^= 1;
		for (int k = 0; k < 4; k++) {
			int tx = nx + d[k][0], ty = ny + d[k][1];
			if (tx < 0 || tx >= R || ty < 0 || ty >= C)
				continue;
			g[tx][ty] ^= 1;
		}
		dfs(nx, ny, s+1);
		g[nx][ny] ^= 1;
		for (int k = 0; k < 4; k++) {
			int tx = nx + d[k][0], ty = ny + d[k][1];
			if (tx < 0 || tx >= R || ty < 0 || ty >= C)
				continue;
			g[tx][ty] ^= 1;
		}
	}
	public int minFlips(int[][] mat) {
		g = mat;
		R = g.length; C = g[0].length;
		dfs(0, -1, 0);
		return min == Integer.MAX_VALUE ? -1 : min;
	}
}

複雜度分析

• 時間複雜度：$O(R × C)$，
• 空間複雜度：$O(R × C)$，

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方法二：dp

狀態壓縮，代辦…

複雜度分析

• 時間複雜度：$O()$，
• 空間複雜度：$O()$，