UVA 11464 Even Parity(枚举)

Description

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 Nintegers (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

题意：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）之和为偶数，如果无解输出-1, 注意到n最大15，第一行只有不超过2^15=32768种可能，接下来根据第一行可以计算出第二行，根据第二行可以计算出第三行，依次类推，总时间复杂度是O（2^n*n^2）

#include <iostream>
#include <string.h>
using namespace std;
const int INF=100000000;
int A[15][15],B[15][15],n;
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;   //sum是B[r-1][c]的上左右3个元素之和
           if(r>1)sum=sum+B[r-2][c];
           if(c>0)sum=sum+B[r-1][c-1];
           if(c<n-1)sum=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()
{
   int T;
   cin>>T;
   for(int M=1;M<=T;M++)
   {
       cin>>n;
       for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
          cin>>A[i][j];
       int mins=INF;
       for(int s=0;s<(1<<n);s++)
        mins=(mins<=check(s)?mins:check(s));
       if(mins==INF)mins=-1;
       cout<<"Case "<<M<<": "<<mins<<endl;
   }
   return 0;
}