UVa 11464 Even Parity (想法题&部分枚举)

11464 - Even Parity

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=456&page=show_problem&problem=2459

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               Output for 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

Case 1: 0  Case 2: 3  Case 3: -1

思路：枚举符合题意的第一行来推出下面的行，然后比较二者是否一样。

复杂度：

完整代码：

/*0.115s*/

#include<cstdio>
#include<cstring>
#include<algorithm>
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 c = 0; c < n; c++)
	{
		if (s & (1 << c)) 
            B[0][c] = 1;
		else if (A[0][c] == 1) 
            return INF; // 1不能变成0
	}
	for (int r = 1; r < n; r++)
		for (int c = 0; c < n; c++)
		{
			int sum = 0; // 元素B[r-1][c]的上、左、右3个元素之和
			if (r > 1) sum += B[r - 2][c];
			if (c > 0) sum += B[r - 1][c - 1];
			if (c < n - 1) sum += B[r - 1][c + 1];
			B[r][c] = sum % 2;
			if (A[r][c] == 1 && B[r][c] == 0) 
                return INF; // 1不能变成0
		}
	int cnt = 0;
	for (int r = 0; r < n; r++)
		for (int c = 0; c < n; c++) 
            if (A[r][c] != B[r][c]) 
                cnt++;
	return cnt;
}

int main(void)
{
	int T;
	scanf("%d", &T);
	for (int kase = 1; kase <= T; kase++)
	{
		scanf("%d", &n);
		for (int r = 0; r < n; r++)
			for (int c = 0; c < n; c++) 
                scanf("%d", &A[r][c]);
		int ans = INF;///设置成无穷大
		for (int s = 0; s < (1 << n); s++)
			ans = min(ans, check(s));
		if (ans == INF) 
            ans = -1;
		printf("Case %d: %d\n", kase, ans);
	}
	return 0;
}