ACM: uva 11464

D

Even Parity Input: Standard Input Output: Standard Output

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
4000
1

1 1 1

0 0 0

Output for Sample Input

Case 1: 0

Case 2: 3

Case 3: -1


题意: n*n的矩阵中,
现在要使得每个格子的上下左右相加的元素和为偶数, 每个格子的元素

         
只能时0和1, 并且只有0元素可以改成1, 问用最少修改次数使得矩阵满足条件.

解题思路:

        
1. 2^(15*15)这是枚举的情况, 可以采用动态规划中状态压缩的思路,
枚举第一行的情况,

            
然后根据第一行的情况推算出其它行的情况. 复杂度O( (2^15)*n*n ).

代码:

#include
<cstdio>

#include <iostream>

#include <cstring>

using namespace std;

#define MAX 16

const int INF = (1<<29);

int n;

int a[MAX][MAX], b[MAX][MAX];

inline int min(int a, int b)

{

    return a
< b ? a: b;

}

int solve(int situation)

{

    memset(b, 0,
sizeof(b));

    for(int i =
0; i < n; ++i)

    {

   
    if(situation
& (1<<i)) b[0][i] =
1;

   
    else
if(a[0][i] == 1) return INF;

    }

   

    for(int i =
1; i < n; ++i)

    {

   
    for(int j =
0; j < n; ++j)

   
    {

   
   
    int sum =
0;

   
   
    if(i
> 1) sum += b[i-2][j];

   
   
    if(j
< n-1) sum += b[i-1][j+1];

   
   
    if(j
> 0) sum += b[i-1][j-1];

   
   
    b[i][j] =
sum%2;

   
   
    if(a[i][j]
== 1 && b[i][j] == 0) return
INF;

   
    }

    }

   

    int count =
0;

    for(int i =
0; i < n; ++i)

   
    for(int j =
0; j < n; ++j)

   
   
    if(a[i][j]
!= b[i][j])

   
   
   
   
count++;

    return
count;

}

int main()

{

//   
freopen("input.txt", "r", stdin);

    int caseNum,
num = 1;

    scanf("%d",
&caseNum);

   
while(caseNum--)

    {

   
    scanf("%d",
&n);

   
    for(int i =
0; i < n; ++i)

   
   
    for(int j =
0; j < n; ++j)

   
   
   
    scanf("%d",
&a[i][j]);

   
   

   
    int result =
INF;

   
    for(int i =
0; i < (1<<n);
++i)

   
    {

   
   
    result =
min(result, solve(i));

   
    }

   
   

   
    if(result ==
INF) printf("Case %d: -1\n", num++);