HDU 5613 Baby Ming and Binary image（扫雷地图的还原）

本文链接：https://blog.youkuaiyun.com/Code_KK/article/details/50574887

Baby Ming and Binary image

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 77 Accepted Submission(s): 31

Problem Description

Baby Ming likes to deal with pixel map in the following way:
First, converting the pixel map into binary image (recorded as 01), and then, in the binary figures, for each grid, add up the grid and its neighboring grids’ binary number, and store in matrix

Mat .
All the pictures Baby Ming chosen have a blank upper margin and bottom margin (the value of which is 0 in the binary image), because he thinks such a picture is beautiful.
The matrix

Mat is so big that Baby Ming is worried about the mistakes in the matrix. So he wants to know whether the binary image can be regained according to the

Mat .

Input

In the first line contains a single positive integer

T , indicating number of test case.

In the second line there are two numbers

n,m , indicating the size of the binary image and the size of the matrix

Mat

In the next

n lines, each line input m numbers, indicating the matrix

Mat .

1≤T≤30,2<n≤12,m≤100

Output

Print Impossible, if it is impossible to regain the binary image.

Print Multiple, if it has more than one result.

Otherwise, print the binary image (Baby Ming's 01 matrix)

Sample Input

Sample Output

  
  
   
   0 0 0 0
0 1 1 1
0 1 0 0
0 0 0 0
Impossible
  
  
  
  

   
   
    
    
     
     Hint
    
    
in the second sample, the matrix[1 0 0] is not Baby Ming's Matrix, so there is no answer

题意：

铭宝宝很喜欢像素图，所以，他喜欢把像素图都做如下的处理：
首先，把像素图变换成二值图（用 $01$ 表示），然后，在二值图中，把每个点以及和它相邻的点（最多 $9$ 个点）的 $01$ 值加起来，并保存在一个矩阵Mat中。
铭宝宝选择的图像底边和顶边都是空白的（二值图中值为 $0$ ），因为他觉得这样的图很漂亮。
不过因为矩阵很大，铭宝宝担心记录过程中出错。所以现在铭宝宝想知道，根据他所保存下来的矩阵Mat，是否能还原出二值图。

题解：枚举最左边一列的所有

2^{n-2}

种情况，然后递推到右边。因为上下两行不用考虑，所以对于每个枚举，递推过程中格子的状态是唯一的，因此对于每一种枚举判断一下是否能构成符合题意的二值图即可。

#include <bits/stdc++.h>
#define LL long long

using namespace std;

int Mat[110][110],Res[110][110],Tmp[110][110];
int n,m;
void Init(int x)
{
    memset(Tmp, 0, sizeof(Tmp));
    for(int i=1; i<=n; i++)
    {
        Tmp[i][1] = (x % 2);
        x >>= 1;
    }
}
bool Judge()
{
    for(int j=1; j<=m; j++)
    {
        if(j == 1 && (Tmp[1][j] || Tmp[n][j]))
            return false;
        for(int i=1; i<=n; i++)
        {
            if(i == 1)
            {
                int num = Mat[i][j] - Tmp[i+1][j-1] - Tmp[i+1][j];
                if(num == 1 || num == 0)
                    Tmp[i+1][j+1] = num;
                else return false;
            }
            else if(i == n - 1)
            {
                int num = Mat[i][j] - (Tmp[i-1][j-1] + Tmp[i-1][j] + Tmp[i-1][j+1] + Tmp[i][j-1] + Tmp[i][j] + Tmp[i][j+1]);
                if(num != 0)
                    return false;
                Tmp[i+1][j+1] = 0;
            }
            else
            {
                int num = Mat[i][j] - (Tmp[i-1][j-1] + Tmp[i-1][j] + Tmp[i-1][j+1] + Tmp[i][j-1] + Tmp[i][j] + Tmp[i][j+1]
                                       +Tmp[i+1][j-1] + Tmp[i+1][j]);
                if(num == 1 || num == 0)
                    Tmp[i+1][j+1] = num;
                else return false;
            }
        }
    }
    for(int i=1; i<=n; i++)
        if(Tmp[i][m+1]) return false;
    return true;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T-- && scanf("%d%d",&n,&m))
    {
        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++)
                scanf("%d",&Mat[i][j]);
        int Cnt = 0;
        for(int i=0; i<(1<<n); i++)
        {
            Init(i);
            if(Judge())
            {
                Cnt++;
                for(int i=1; i<=n; i++)
                    for(int j=1; j<=m; j++)
                        Res[i][j] = Tmp[i][j];
            }
        }
        if(Cnt == 1)
        {
            for(int i=1; i<=n; i++)
                for(int j=1; j<=m; j++)
                    printf(j == m ? "%d\n":"%d ",Res[i][j]);
        }
        else if(Cnt < 1)
            puts("Impossible");
        else
            puts("Multiple");
    }
    return 0;
}