UVA 11464 Even Parity（部分列舉 遞推）

賈樹丙發表於2013-07-18

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

 

題目大意：給你一個n*n的01矩陣（每個元素非0即1），你的任務是把儘量少的0變成1，使得每個元素的上、下、左、右的元素（如果存在的話）之和均為偶數。

 

分析：偶數矩陣，關燈遊戲改版。直接列舉會超時。注意到n只有15，第一行只有不超過2^15=32768中可能，所以第一行的情況可以列舉。接下來根據第一行可以完全計算出第2行，根據第二行又能計算出第三行...

 

程式碼如下：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 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   memset(B, 0, sizeof(B));
12   for(int c = 0; c < n; c++) {
13     if(s & (1<<c)) B[0][c] = 1;
14     else if(A[0][c] == 1) return INF; // 1不能變成0
15   }
16   for(int r = 1; r < n; r++)
17     for(int c = 0; c < n; c++) {
18       int sum = 0; // 元素B[r-1][c]的上、左、右3個元素之和
19       if(r > 1) sum += B[r-2][c];
20       if(c > 0) sum += B[r-1][c-1];
21       if(c < n-1) sum += B[r-1][c+1];
22       B[r][c] = sum % 2;
23       if(A[r][c] == 1 && B[r][c] == 0) return INF; // 1不能變成0
24     }
25   int cnt = 0;
26   for(int r = 0; r < n; r++)
27     for(int c = 0; c < n; c++) if(A[r][c] != B[r][c]) cnt++;
28   return cnt;
29 }
30 
31 int main() {
32   int T;
33   scanf("%d", &T);
34   for(int kase = 1; kase <= T; kase++) {
35     scanf("%d", &n);
36     for(int r = 0; r < n; r++)
37       for(int c = 0; c < n; c++) scanf("%d", &A[r][c]);
38 
39     int ans = INF;
40     for(int s = 0; s < (1<<n); s++)
41       ans = min(ans, check(s));
42     if(ans == INF) ans = -1;
43     printf("Case %d: %d\n", kase, ans);
44   }
45   return 0;
46 }

 

     

相關文章