Uva11464 - Even Parity

最新推荐文章于 2020-09-26 12:29:30 发布

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本文探讨了一个特定的编程问题：在一个N×N的网格中，通过将部分0转换为1，使得每个单元格的奇偶性变为偶数。文章提供了一段示例代码，详细解释了如何实现这一目标并计算所需的最小转换次数。

We have a grid of size N 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 4:
1 0 1 0 The parity of each cell would be 1 3 1 2
1 1 1 1 2 3 2 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 rst 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
Sample Output
Case 1: 0
Case 2: 3
Case 3: -1

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <queue>
using namespace std;
const int MAXN = 20;
const int MAX = 1<<20;
int n, ans;
int mp[MAXN][MAXN];
struct node{
    int x, y;
}q[MAXN*MAXN];
int top;
void dfs(int xb,int select)
{
    if(xb==n+1)
    {
        int temp = select;
        bool pd = true;
        top = 0;
        for(int i=1; pd && i<n; i++)
        {
            for(int j=1; pd && j<=n; j++)
            {
                int s = mp[i-1][j] + mp[i+1][j] + mp[i][j+1] + mp[i][j-1];
                if(s%2)
                {   
                    if(mp[i+1][j]==1)
                    {
                        pd = false;
                    }else{
                       q[top].x = i+1;
                       q[top++].y = j;
                        mp[i+1][j]=1;
                        temp++;
                    }
                }
            }
        }
        for(int j=1;  pd &&j<=n; j++)
        {
            int s = mp[n-1][j] + mp[n+1][j] + mp[n][j+1] + mp[n][j-1];
            if(s%2)
            {
                 pd = false;
            }
        }
        if(pd && ans>temp)  ans=temp;
        for(int i=0; i<top; i++)
        {
            mp[q[i].x][q[i].y] = 0;
        }
        return;
    }
    if(mp[1][xb]==1)
    {
        dfs(xb+1,select);
    }else{
        dfs(xb+1,select);
        mp[1][xb] = 1;
        dfs(xb+1,select+1);
        mp[1][xb] = 0;
    }
}
int main()
{
    int t, cs, i, j;
    scanf("%d",&t);
    for(cs=1; cs<=t; cs++)
    {
        scanf("%d",&n);
        memset(mp,0,sizeof (mp));
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=n; j++)
            {
                scanf("%d",&mp[i][j]);
            }
        }
        ans = MAX;
        dfs(1,0);
        if(ans==MAX) ans=-1;
        printf("Case %d: %d\n",cs,ans);
    }
    return 0;
}