知识点：递归，递推

和李煜东的那道例题大体思想上一样，指数枚举第一行的时间复杂度是可以接受的，所以枚举第一行，然后由第一行递推得出后面的行，在中间判断存结果就行，需要注意的是，本题只能0变成1，这个不要想当然我一开始做的时候直接默认01可以互换

#include <bits/stdc++.h>

#define fi first
#define se second
#define pb push_back
#define mk make_pair
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()

using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pa;

const int Inf = 1000000000;

int n, mat[20][20], t[20][20], ans;
int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};
vi chosen;

void calc(int xx) {
    if (xx == n + 1) {
        int cnt = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) t[i][j] = mat[i][j];
        }
        for (int i = 0; i < sz(chosen); i++) {
            if (t[1][chosen[i]]) continue;
            t[1][chosen[i]] = 1;
            cnt++;
        }
        for (int i = 2; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                int temp = t[i - 2][j] + t[i][j] + t[i - 1][j - 1] + t[i - 1][j + 1];
                if (temp % 2) {
                    if (t[i][j]) return;
                    cnt++; t[i][j] = 1;
                }
                if (cnt >= ans) return;
            }
        }
        int ok = 1;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                int temp = t[i - 1][j] + t[i + 1][j] + t[i][j - 1] + t[i][j + 1];
                if (temp % 2) ok = 0;
            }
        }
        if (ok && cnt < ans) ans = cnt;
        return;
    }
    calc(xx + 1);
    chosen.pb(xx);
    calc(xx + 1);
    chosen.pop_back();
}

int main() {
    int t;
    cin >> t;
    for (int k = 1; k <= t; k++) {
        cin >> n;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) cin >> mat[i][j];
        }
        ans = Inf;
        calc(1);
        cout << "Case " << k << ": ";
        cout << (ans == Inf ? -1 : ans) << endl;
    }
    return 0;
}