116. 飛行員兄弟

題目連結：

https://www.acwing.com/problem/content/118/

題解：

1、遞迴：

遞迴的最難理解的點就是要從滿足題目的從左向右，從上到下，所以遇到y == 4的邊界時，就應該跳到下一行的第一個位置，注意恢復現場。

2、位運算：

和95、費解的開關差不多，但是這裡是窮舉2^16種可能，直接判斷，利用getValue(int x,int y)將物理一維轉化為二維的窮舉。

AC程式碼：

1、遞迴

public class NO116 {
    static char[][] arr = new char[4][4];
    static LinkedList<Pair> res = new LinkedList<>();
    static Pair[] pairRes = null;
    static int step = 1000;

    static void turn(int x,int y) {
        for (int i = 0; i < 4; i++) {
            arr[x][i] ^= 1;
            arr[i][y] ^= 1;
        }
        arr[x][y] ^= 1;
    }

    static void dfs(int x,int y ) {
        if (x == 3 && y == 4) {
            boolean flag = true;
            pos:
            for (int i = 0 ;i < 4;i++) {
                for (int j = 0;j<4;j++) {
                    if (arr[i][j] == ‘0‘){
                        flag = false;
                        break pos;
                    }
                }
            }
            if (flag && res.size() < step) {
                step = res.size();
                pairRes = res.toArray(new Pair[0]);
            }
            return;
        }
        if (y == 4) { // 從左向右，從上到下
            x = x+1;
            y = 0;
        }
        // 操作門把手
        turn(x,y);
        res.push(new Pair(x+1,y+1));
        dfs(x,y+1);
        // 恢復現場
        turn(x,y);
        res.pop();
        // 不操作門把手
        dfs(x,y+1);
    }
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        for (int i = 0; i < 4; i++) {
            arr[i] = sc.next().toCharArray();
        }
        for (int i = 0; i < 4; i++) {
            for (int j = 0; j < 4; j++) arr[i][j] = arr[i][j] == ‘-‘ ? ‘1‘ : ‘0‘;
        }
        dfs(0,0);
        System.out.println(step);
        for (int i = pairRes.length-1; i >= 0; i--) {
            System.out.println(pairRes[i].x + " " + pairRes[i].y);
        }
    }
}

class Pair {
    int x,y;
    public Pair() {}
    public Pair(int x,int y) {
        this.x = x;
        this.y = y;
    }
}
 

2、位運算

public class NO116 {
    static char[][] arr = new char[4][4];
    static char[][] temp = new char[4][4];
    static List<Pair> res = new ArrayList<>();
    static Pair[] pairRes = null;
    static int step = 1000;

    static int getValue(int x,int y) {
        return x*4 + y;
    }

    static void turn(int x,int y) {
        for (int i = 0; i < 4; i++) {
            arr[x][i] ^= 1;
            arr[i][y] ^= 1;
        }
        arr[x][y] ^= 1;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        for (int i = 0; i < 4; i++) {
            arr[i] = sc.next().toCharArray();
        }
        for (int i = 0; i < 4; i++) {
            for (int j = 0; j < 4; j++) arr[i][j] = arr[i][j] == ‘-‘ ? ‘1‘ : ‘0‘;
        }
        for (int op = 0;op < 1<<16;op++) {
            for (int i = 0;i <4;i++) temp[i] = Arrays.copyOf(arr[i],arr[i].length);
            res.clear();
            for (int i = 0;i < 4;i++) {
                for (int j = 0;j < 4;j++) {
                    if( (op >> getValue(i,j) & 1) == 1) {
                        turn(i,j);
                        res.add(new Pair(i+1,j+1));
                    }
                }
            }
            boolean flag = true;
            pos:
            for (int i = 0; i < 4; i++) {
                for (int j = 0; j < 4; j++) {
                    if (arr[i][j] == ‘0‘) {
                        flag = false;
                        break pos;
                    }
                }
            }
            if (flag && res.size() < step) {
                step = res.size();
                pairRes = res.toArray(new Pair[0]);
            }
            for (int i = 0;i <4;i++) arr[i] = Arrays.copyOf(temp[i],temp[i].length);
        }


        System.out.println(step);
        for (int i = 0; i < pairRes.length; i++) {
            System.out.println(pairRes[i].x + " " + pairRes[i].y);
        }
    }
}

class Pair {
    int x,int y) {
        this.x = x;
        this.y = y;
    }
}