题解思路来源于：费解的开关 https://www.acwing.com/file_system/file/content/whole/index/content/3646/

相关题解： https://acwing.com/file_system/file/content/whole/index/content/12664204/

枚举16个开关的所有状态(用二进制数来表示)求解最小值即可

import java.io.*;
import java.util.*;

public class Main {
    static char[][] g = new char[6][6];
    static ArrayList<int[]> ans = new ArrayList<>();

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        for (char i = 1; i <= 4; i++) g[i] = toArr(sc.nextLine().toCharArray());

        int cnt = work();
        //输出结果
        out.println(cnt);
        for (int[] item : ans) {
            out.println(item[0] + " " + item[1]);
        }

        out.flush();
    }
    /**
    * 索引从1开始的
    * 返回索引u对应的x坐标(一维转二维)
    */
    static int getX(int u) {
        return (u - 1) / 4 + 1;
    }
    /**
    * 返回索引u对应的y坐标(一维转二维)
    */
    static int getY(int u) {
        return (u - 1) % 4 + 1;
    }

    static void turn(int x, int y) {
        for (int i = 1; i <= 4; i++) {
            if (g[x][i] == '-') g[x][i] = '+';
            else g[x][i] = '-';
            if (g[i][y] == '-') g[i][y] = '+';
            else g[i][y] = '-';
        }
        if (g[x][y] == '-') g[x][y] = '+';
        else g[x][y] = '-';
    }

    static int work() {
        int cnt = 1000000;

        for (int k = 0; k < 1 << 16; k++) {
            ArrayList<int[]> list = new ArrayList<>();
            int res = 0;
            char[][] backup = new char[6][6];
            for (int i = 1; i <= 4; i++) backup[i] = Arrays.copyOf(g[i], 6);

            for (int i = 0; i < 16; i++) {
                if ((k >> i & 1) == 1) {
                    res++;
                    //由于我的索引从1开始，所以有1的偏移量
                    int x = getX(i + 1), y = getY(i + 1);
                    list.add(new int[]{x, y});
                    turn(x, y);
                }
            }

            boolean flag = true;
            for (int i = 1; i <= 4; i++) {
                boolean flag2 = false;
                for (int j = 1; j <= 4; j++)
                    if (g[i][j] == '+') {
                        flag = false;
                        flag2 = true;
                        break;
                    }
                if (flag2) break;
            }
            //不添加等于就符合了题目的要求了
            //该枚举方式本来就是自上到下，从左到右
            if (flag && res < cnt) {
                cnt = res;
                ans.clear();
                ans.addAll(list);
            }

            for (int i = 1; i <= 4; i++) g[i] = Arrays.copyOf(backup[i], 6);
        }

        return cnt;
    }

    static char[] toArr(char[] str) {
        char[] res = new char[6];
        for (int i = 1; i <= 4; i++) res[i] = str[i - 1];
        return res;
    }
}

算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

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算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla