UVa 11464 Even Parity

　　对所有可能进行枚举，不过情况太多了，只需枚举第一行的情况，剩下来的就可以确定了。

　　代码如下：

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int maxn = 20;
const int INF = 1000000000;
int n, a[maxn][maxn], b[maxn][maxn];

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

int main()
{
#ifdef LOCAL
    freopen("in", "r", stdin);
#endif
    int T;
    scanf("%d", &T);
    for(int kase = 1; kase <= T; kase++)
    {
        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\n", kase, ans);
    }
    return 0;
}