uva 11464 - Even Parity（3级）

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                             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

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Problem Setter: Sohel Hafiz,

Special Thanks: Derek Kisman, Md. Arifuzzaman Arif

 

思路：状态压缩枚举第一行的状态，然后根据前面的状态推出后面的状态。然后用初状态与目标状态比较。

 

#include<iostream>
#include<cstdio>
#include<cstring>
#define FOR(i,n) for(int i=0;i<n;++i)
using namespace std;
const int mm=17;
const int oo=1e9;
int g[mm][mm],t[mm][mm];
int n;
void check(int z,int&ans)
{ memset(t,0,sizeof(t));///目标状态
  FOR(i,n)
  if(z&(1<<i))t[0][i]=1;
  else if(g[0][i])return;///1无法变为0
  int num;
  for(int i=1;i<n;++i)FOR(j,n)
  {num=0;
    if(i>1)num+=t[i-2][j];///up
    if(j)num+=t[i-1][j-1];///left
    if(j<n-1)num+=t[i-1][j+1];
    if(num&1){t[i][j]=1;continue;}
      if(g[i][j])return;
  }
  int ret=0;
  FOR(i,n)FOR(j,n)
  if(g[i][j]^t[i][j])
    ++ret;
  ans=ans<ret?ans:ret;
}
int main()
{
  int cas;
  while(~scanf("%d",&cas))
  {
    FOR(ca,cas)
    {
      scanf("%d",&n);
      FOR(i,n)FOR(j,n)
      scanf("%d",&g[i][j]);
      int ans=oo;
      FOR(i,1<<n)
      check(i,ans);
      if(ans==oo)
      ans=-1;
      printf("Case %d: %d\n",ca+1,ans);
    }
  }
}