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

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

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

     int ret = ;
     for (int i=; i<n; ++i)    {
         for (int j=; 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 = ;  scanf ("%d", &t);
     while (t--) {
         scanf ("%d", &n);
         for (int i=; i<n; ++i)    {
             for (int j=; j<n; ++j)    {
                 scanf ("%d", &a[i][j]);
             }
         }

         int ans = INF;
         for (int i=; i<(<<n); ++i)    {
             ans = min (ans, check (i));
         }

         if (ans == INF) ans = -;
         printf ("Case %d: %d\n", ++cas, ans);
     }

     return ;
 }