UVA 11464 Even Parity（部分枚举 递推）

Even Parity

We have a grid of size N x N. Each cell of the grid initially contains a zero(0) or a one(1). 
The parity of a cell is the number of 1s surrounding that cell. A cell is surrounded by at most 4 cells (top, bottom, left, right).

Suppose we have a grid of size 4 x 4:

1	0	1	0	The parity of each cell would be	1	3	1	2
1	1	1	1		2	3	3	1
0	1	0	0		2	1	2	1
0	0	0	0		0	1	0	0

For this problem, you have to change some of the 0s to 1s so that the parity of every cell becomes even. We are interested in the minimum number of transformations of 0 to 1 that is needed to achieve the desired requirement.

Input

The first line of input is an integer T (T<30) that indicates the number of test cases. Each case starts with a positive integer N(1≤N≤15). Each of the next N lines contain N integers (0/1) each. The integers are separated by a single space character.

Output

For each case, output the case number followed by the minimum number of transformations required. If it's impossible to achieve the desired result, then output -1 instead.

Sample Input

3

3

0 0 0

0 0 0

0 0 0

3

0 0 0

1 0 0

0 0 0

3

1 1 1

1 1 1

0 0 0

Output for Sample Input

Case 1: 0 
Case 2: 3 
Case 3: -1

题目大意：给你一个n*n的01矩阵（每个元素非0即1），你的任务是把尽量少的0变成1，使得每个元素的上、下、左、右的元素（如果存在的话）之和均为偶数。

分析：偶数矩阵，关灯游戏改版。直接枚举会超时。注意到n只有15，第一行只有不超过2^15=32768中可能，所以第一行的情况可以枚举。接下来根据第一行可以完全计算出第2行，根据第二行又能计算出第三行...

代码如下：

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 
 const int maxn = ;
 const int INF = ;
 int n, A[maxn][maxn], B[maxn][maxn];
 
 int check(int s) {
   memset(B, , sizeof(B));
   for(int c = ; c < n; c++) {
     if(s & (<<c)) B[][c] = ;
     else if(A[][c] == ) return INF; // 1不能变成0
   }
   for(int r = ; r < n; r++)
     for(int c = ; c < n; c++) {
       int sum = ; // 元素B[r-1][c]的上、左、右3个元素之和
       if(r > ) sum += B[r-][c];
       if(c > ) sum += B[r-][c-];
       if(c < n-) sum += B[r-][c+];
       B[r][c] = sum % ;
       if(A[r][c] ==  && B[r][c] == ) return INF; // 1不能变成0
     }
   int cnt = ;
   for(int r = ; r < n; r++)
     for(int c = ; c < n; c++) if(A[r][c] != B[r][c]) cnt++;
   return cnt;
 }
 
 int main() {
   int T;
   scanf("%d", &T);
   for(int kase = ; kase <= T; kase++) {
     scanf("%d", &n);
     for(int r = ; r < n; r++)
       for(int c = ; c < n; c++) scanf("%d", &A[r][c]);
 
     int ans = INF;
     for(int s = ; s < (<<n); s++)
       ans = min(ans, check(s));
     if(ans == INF) ans = -;
     printf("Case %d: %d\n", kase, ans);
   }
   return ;
 }