UVA11464 BNU19882 Even Parity

Even Parity

Time Limit: 3000ms

Memory Limit: 131072KB

This problem will be judged on  UVA. Original ID:  11464
64-bit integer IO format:  %lld      Java class name:  Main

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

https://www.bnuoj.com/v3/problem_show.php?pid=19882

给你一个01矩阵(每个元素非0即1),你的任务是把尽量少的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.

如果你枚举每个数字情况太多 只需要枚举第一行其他行就可以确定了

#include<stdio.h>
#include<string.h>
using namespace std;
int ap1[20][20],ap2[20][20];
#define inf 0x3f3f3f3f
int main()
{
    int t,n,h,check,k,i,j,MIN,sum,u;
    scanf("%d",&t);
    for(int kk=1; kk<=t; kk++)
    {
        scanf("%d",&n);
        for(i=1; i<=n; i++)
            for(j=1; j<=n; j++)
                scanf("%d",&ap1[i][j]);
        printf("Case %d: ",kk);
        memset(ap2,0,sizeof(ap2));
        int N=1<<n;
        N--;
        //printf("%d\n",N);
        MIN=inf;  //初始化
        for(h=0; h<=N; h++)
        {
            for(k=1; k<=n; k++)
                ap2[1][k]=(1&(h>>(k-1)));
            for(i=2; i<=n; i++)
                for(j=1; j<=n; j++)
                    ap2[i][j]=(ap2[i-1][j-1]+ap2[i-1][j+1]+ap2[i-2][j])%2==0?0:1;
            int gg=0;
            for(i=1; i<=n; i++)     //判断矩形成立吗
            {
                for(j=1; j<=n; j++)
                    if((ap2[i-1][j]+ap2[i+1][j]+ap2[i][j-1]+ap2[i][j+1])%2)
                    {
                        gg=1;
                        break;
                    }
                if(gg)
                    break;
            }
            check=0;
            sum=0;
            if(gg) continue;
            for(i=1; i<=n; i++)
            {
                for(j=1; j<=n; j++)
                    if(ap1[i][j]!=ap2[i][j])
                    {
                        if(ap2[i][j]==0)
                        {
                            check=1;
                            sum=inf;
                            break;
                        }
                        else sum++;
                    }

                if(check) break;
            }
            if(sum<MIN)
            {
                MIN=sum;
            }
        }
        if(MIN==inf) printf("-1\n");
        else printf("%d\n",MIN);
    }
    return 0;
}