MatrixPainting

MatrixPainting
Problem Statement

There is a matrix with 9 rows and 9 columns. Each cell of the matrix is either black or white. With a single repaint operation, you can repaint all the cells in a single row or column black if the row or column already contains at least 5 black cells. Your goal is to make all the cells in the matrix black using a minimal number of repaint operations.

You will be given a String[] matrix, where the jth character of the ith element represents the cell at row i, column j. Black cells will be written as '1' (one), and white cells will be written as '0' (zero). Return the minimal number of repaint operations required to make all the cells black, or -1 if this is impossible.
 
Definition
       
Class:    MatrixPainting
Method:    countRepaints
Parameters:    String[]
Returns:    int
Method signature:    int countRepaints(String[] matrix)
(be sure your method is public)
   
 
Constraints
-    matrix will contain exactly 9 elements.
-    Each element of matrix will contain exactly 9 characters.
-    Each element of matrix will consist of '0' and '1' characters only.
 
Examples
0)   
{"001111111",
 "011111111",
 "011111111",
 "011111111",
 "011111111",
 "101111111",
 "101111111",
 "101111111",
 "101111111"}

Returns: 3

First, you should repaint the first row. After that, you can repaint the first and the second column.

总共有9行9列，共2¹⁸种repaint的方法。现在的任务是判断每种repaint是否能够满足要求

int valid(const vector<int>& way, vector<string>& matrix)
    ...{
        int i,k,a,b;
        int sum,all;
        all = 0;
        for(a=0;a<9;a++)
        ...{
            for(b=0;b<9;b++)
            ...{
                if(matrix[a][b] == '1') all += 1;
            }
        }
        bool find;
        int mm = 0;
        if(all == 81) return 0;
        while(true)
        ...{
            find = false;
            mm ++;
            for(i=0;i<way.size();i++)
            ...{
                k = way[i]%9;
                if(way[i]>9)//column
                ...{
                    sum = 0;
                    for(a = 0; a < 9; a++)
                    ...{
                        if(matrix[a][k] == '1') sum += 1;
                    }
                    if(sum >= 5 && sum < 9)
                    ...{
                        find = true;
                        for(a = 0;a < 9; a++)
                        ...{
                            if(matrix[a][k] == '0')
                            ...{
                                matrix[a][k] = '1';
                                all ++;
                            }
                        }
                    }
                    if(all == 81) return mm;
                }
                else
                ...{
                    sum = 0;
                    for(a = 0; a < 9; a++)
                    ...{
                        if(matrix[k][a] == '1') sum += 1;
                    }
                    if(sum >= 5 && sum < 9)
                    ...{
                        find = true;
                        for(a = 0;a < 9; a++)
                        ...{
                            if(matrix[k][a] == '0')
                            ...{
                                matrix[k][a] = '1';
                                all ++;
                            }
                        }
                    }
                    if(all == 81) return mm;
                }
                if(find)
                ...{
                    break;
                }
            }
            //cout<<"i"<<i<<"  "<<mm<<" "<<all<<endl;
            if(!find) return -1;
        }
    }

还需要一个函数生成所有可能的repaint行和列

    int allpaints(vector< vector<int> > & output)
    ...{
        int i,j,size;
        vector<int> oneway;
        output.push_back(oneway);
        
        for(i=1;i<=18;i++)
        ...{
            vector<int> way;
            size = output.size();
            for(j=0;j<size;j++)
            ...{
                way = output[j];
                output[j].push_back(i);
                output.push_back(way);
            }
        }
        return 0;
    }

最终的函数实现如下：

    int countRepaints(vector<string> matrix)
    ...{
        vector<vector<int> > path;
        allpaints(path);
        int i;
        int size = 19;
        int mm;
        for(i=0;i<path.size();i++)
        ...{
            if(path[i].size() >= size) continue;
            vector<string> M = matrix;
             mm = valid(path[i],M);
             //cout<<"mm:"<<mm<<endl;
            if(mm >= 0)
            ...{
                if(size > mm) size = mm;
                //cout<<size<<endl;
            }
        }
        if(size == 19) return -1;
        return size;
    }