【深搜+回溯+枚举】poj 755 Flip Game

题目见：http://noi.openjudge.cn/ch0201/755/

最多翻转15次，所以按次数遍历，对0，1，2…15次中的每一次调用play()进行枚举，注意函数的返回条件以及回溯的处理。

#include 
#include 
using namespace std;

char a[8][8];
int step;
bool flag;

bool isOver() {
    char temp = a[0][0];
    for (int i = 0; i < 4; i++) {
        for (int j = 0; j < 4; j++) {
            if (a[i][j] != temp)
                return false;
        }
    }
    return true;
}

void change(int m, int n) {
    if (m < 0 || m > 3 || n < 0 || n > 3) return;
    if (a[m][n] == 'b') a[m][n] = 'w';
    else
        a[m][n] = 'b';
    return;
}

void flip(int m, int  n) {
    change(m, n);
    change(m - 1, n);
    change(m + 1, n);
    change(m, n - 1);
    change(m, n + 1);
    return;
}

void play(int m, int n, int deep) {
    if (deep == step) {
        flag = isOver();
        return;
    }
    if (flag || m > 3)
        return;
    flip(m, n);
    if (n < 3)
        play(m, n + 1, deep + 1);
    else
        play(m + 1, 0, deep + 1);
    flip(m, n);
    if (n < 3)
        play(m, n + 1, deep);
    else
        play(m + 1, 0, deep);
    return;
}

int main() {
    for (int i = 0; i < 4; i++)
        for (int j = 0; j < 4; j++)
            cin >> a[i][j];
    flag = false;
    for (step = 0; step < 16; step++) {
        play(0, 0, 0);
        if (flag == true)
            break;
    }
    if (flag == true)
        cout << step << endl;
    else
        cout << "Impossible" << endl;
    //system("pause");
    return 0;
}