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：

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 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <cstring>
 4 using namespace std;
 5 
 6 const int maxn = 20;
 7 const int INF = 1000000000;
 8 int n, a[maxn][maxn], b[maxn][maxn];
 9 
10 int check(int s)
11 {
12     memset(b, 0, sizeof(b));
13     for(int i = 0; i < n; i++)
14     {
15         if(s & (1<<i))   b[0][i] = 1;
16         else if(a[0][i] == 1)   return INF;
17     }
18     for(int i = 1; i < n; i++)
19         for(int j = 0; j < n; j++)
20         {
21             int sum = 0;
22             if(i > 1)   sum += b[i-2][j];
23             if(j > 0)   sum += b[i-1][j-1];
24             if(j < n-1)   sum += b[i-1][j+1];
25             b[i][j] = sum % 2;
26             if(a[i][j] == 1 && b[i][j] == 0)   return INF;
27         }
28     int cnt = 0;
29     for(int i = 0; i < n; i++)
30         for(int j = 0; j < n; j++)
31             if(a[i][j] != b[i][j])   cnt++;
32     return cnt;
33 }
34 
35 int main()
36 {
37 #ifdef LOCAL
38     freopen("in", "r", stdin);
39 #endif
40     int T;
41     scanf("%d", &T);
42     for(int kase = 1; kase <= T; kase++)
43     {
44         scanf("%d", &n);
45         for(int i = 0; i < n; i++)
46             for(int j = 0; j < n; j++)
47                 scanf("%d", &a[i][j]);
48         int ans = INF;
49         for(int i = 0; i < (1<<n); i++)
50             ans = min(ans, check(i));
51         if(ans == INF) ans = -1;
52         printf("Case %d: %d\n", kase, ans);
53     }
54     return 0;
55 }