UVa 11464 偶数矩阵

题意：

给出一个01矩阵，把尽量少的0变成1，使得每个元素的上下左右的元素之和均为偶数。

思路：
用二进制枚举第一行的情况，之后每一行都可以由上一行推导得出。

 1 #include<iostream> 
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<string>
 5 using namespace std;
 6 
 7 #define INF 10000000
 8 int g[20][20];
 9 int G[20][20];
10 int N;
11 
12 int check(int s) 
13 {
14     memset(G, 0, sizeof(G));
15     for (int i = 0; i < N; i++){
16         if (s & (1 << i))  G[0][i] = 1;
17         else if (g[0][i] == 1) return INF;
18     }
19     for (int i = 1; i < N; i++) 
20     {
21         for (int j = 0; j < N; j++)
22         {
23             int sum = 0;
24             if (i > 1) sum += G[i - 2][j];
25             if (j > 0) sum += G[i - 1][j - 1];
26             if (j < N - 1) sum += G[i - 1][j + 1];
27             G[i][j] = sum % 2;
28             if (g[i][j] == 1 && G[i][j] == 0) return INF;
29         }
30     }
31     int cnt = 0;
32     for (int i = 0; i <N; i++)
33     for (int j = 0; j < N; j++) {
34         if (G[i][j] != g[i][j])
35             cnt++;
36     }
37     return cnt;
38 }
39 int main() 
40 {
41     //freopen("D:\txt.txt", "r", stdin);
42     int T;
43     cin >> T;
44     for (int kase = 1; kase <= T;kase++)
45     {
46         scanf("%d", &N);
47         for (int i = 0; i < N; i++)
48         for (int j = 0; j < N; j++)
49                 scanf("%d", &g[i][j]);
50         int ans = INF;
51 
52         for (int i = 0; i < (1 << N); i++) {
53             ans = min(ans, check(i));
54         }
55         printf("Case %d: ", kase);
56         if (ans == INF) printf("-1
");
57         else printf("%d
", ans);
58     }
59     return 0;
60 }