状态压缩+枚举 UVA 11464 Even Parity

题目传送门

/*
    题意：求最少改变多少个0成1，使得每一个元素四周的和为偶数
　　状态压缩+枚举：枚举第一行的所有可能(1<<n)，下一行完全能够由上一行递推出来，b数组保存该位置需要填什么
     　　　　　　　　最后检查不同的数量，取最小值
*/
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int MAXN = 20;
const int INF = 0x3f3f3f3f;
int a[MAXN][MAXN], b[MAXN][MAXN];
int n;

int check(int s) {
    memset (b, 0, sizeof (b));
    for (int i=0; i<n; ++i)    {
        if (s & (1 << i))   b[0][i] = 1;
        else if (a[0][i] == 1)  return INF;
    }

    for (int i=1; i<n; ++i)    {
        for (int j=0; j<n; ++j)    {
            int sum = 0;
            if (i > 1)  sum += b[i-2][j];
            if (j > 0)  sum += b[i-1][j-1];
            if (j < n-1)  sum += b[i-1][j+1];
            b[i][j] = sum % 2;
            if (a[i][j] == 1 && b[i][j] == 0)   return INF;
        }
    }

    int ret = 0;
    for (int i=0; i<n; ++i)    {
        for (int j=0; j<n; ++j)    {
            if (a[i][j] != b[i][j]) ret++;
        }
    }

    return ret;
}

int main(void)  {       //UVA 11464 Even Parity
    //freopen ("G.in", "r", stdin);
    
    int t, cas = 0;  scanf ("%d", &t);
    while (t--) {
        scanf ("%d", &n);
        for (int i=0; i<n; ++i)    {
            for (int j=0; j<n; ++j)    {
                scanf ("%d", &a[i][j]);
            }
        }
        
        int ans = INF;
        for (int i=0; i<(1<<n); ++i)    {
            ans = min (ans, check (i));
        }
        
        if (ans == INF) ans = -1;
        printf ("Case %d: %d
", ++cas, ans);
    }

    return 0;
}