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【原创】一个基于简单剪枝的DFS解数独程序
问题来源:leetCode Sudoku Solver
Write a program to solve aSudoku puzzle by filling the empty cells.
Empty cells are indicated by the character *.*.
You may assume that there will be only one unique solution.
问题链接:
https://oj.leetcode.com/problems/sudoku-solver/
功能:
自动解数独题
大概方案:
采用以行为单位的DFS来寻找答案。
提纲:
0、对全部数据进行候选数预处理,记录下每个空格的候选数。
1、根据候选数表和当前结果表,以行为单位,递归产生所有可能的行,并加入结果数组,然后对每一个产生的可能行进行新一轮的递归调用。这里有一个互相递归调用的行为。
2、若无法产生可能的行或产生完所有行的下一级返回假,则返回假。
3、若有一个下级递归返回真或者最后一行中成功产生了可能的行,返回真。
主要方法概览:
(注意:候选数表为全局数组,所有方法可以直接访问)
0、bool tryRow(introw,vector<vector<int> > &res);
该方法为对某行进行数字试填,row为某行的行号,res为当前结果数组。
1、bool fillRow(introw,int col,vector<int> &now,vector<vector<int> >&res);
该方法为对特定的格子进行数字试填,row为行号,col为列号,now为当前行中已填的数字,res为结果数组。
2、以上两种方法互相递归调用。对于特定新行,调用tryRow方法,而填新行内的数则递归调用fillRow,当当前行已经填满后,在fillRow中调用新一轮的tryRow方法进入下一行。
Solution类的全部代码和注释:(\t被吞,用/*...*/来缩进)
#include <iostream>
#include <vector>
#include <cstdio>
using namespace std;
class Solution
{
public:
/*......*/ int sel[9][9][10];
// 候选数数组,sel[x][y][0]标记横坐标为x,纵坐标为y的格子是否有数字,有则为该数字,
// 待填则sel[x][y][num](1<= num <=9)为1表示该格子可选数包括num.
/*......*/ Solution()
/*......*/ {
/*.........*/ for (int i = 0;i < 9;i++)
/*............*/ for (int j = 0;j <9;j++)
/*...............*/ for (int p = 0;p <10;p++)
/*..................*/ sel[i][j][p] = 1;
/*......*/ }
/*......*/ voidsolveSudoku(vector<vector<char> > &board) // 主处理方法
/*......*/ {
/*.........*/ vector<vector<int>> res;
/*.........*/ init(board); // 进行候选数初始化处理
/*.........*/ tryRow(0,res);
/*.........*/ int len1 = board.size();
/*.........*/ int len2 = board[0].size();
/*.........*/ for (int i = 0;i <len1;i++)
/*............*/ for (int j = 0;j <len2;j++)
/*...............*/ if (board[i][j] == *.*)
/*..................*/ board[i][j] =res[i][j] + *0*;
/*......*/ }
/*......*/ bool tryRow(introw,vector<vector<int> > &res)
/*......*/ {
/*.........*/ if (row >= 9)
/*.........*/ {
/*............*/ return true;
/*.........*/ }
/*.........*/ bool flag = false;
/*.........*/ vector<int> now(10);
/*.........*/ flag =fillRow(row,0,now,res);
/*.........*/ return flag;
/*......*/ }
/*......*/ bool fillRow(int row,intcol,vector<int> &now,vector<vector<int> > &res)
/*......*/ {
/*.........*/ bool flag = false;
/*.........*/ if (col >= 9)
/*.........*/ {
/*............*/ res.push_back(now);
/*............*/ flag = tryRow(row +1,res);
/*............*/ if (flag == false)
/*...............*/ res.pop_back();
/*............*/ return flag;
/*.........*/ }
/*.........*/ if (sel[row][col][0] != 0)
/*.........*/ {
/*...............*/ now[col] =sel[row][col][0];
/*...............*/ flag = fillRow(row,col+ 1,now,res);
/*...............*/ return flag;
/*.........*/ }
/*.........*/ int hash[10] = {0}; // now中现存的数为1,该数组中为1的数不能选
/*.........*/ for (int i = col - 1;i >=0;i--) // 判断当前行中需要更新的候选数
/*............*/ hash[now[i]] = 1;
/*.........*/ update(row,col,hash,res); // 该函数继续更新该格子的候选数
/*......*/ for (int i = 1;i <= 9;i++)
/*.........*/ {
/*............*/ if (hash[i] == 0&& sel[row][col][i] == 1)
/*............*/ {
/*...............*/ now[col] = i;
/*...............*/ flag = fillRow(row,col+ 1,now,res);
/*...............*/ if (flag)
/*..................*/ return true;
/*............*/ }
/*.........*/ }/*...*/
/*.........*/ return flag;
/*......*/ }/*...*/
/*......*/
/*......*/ void update(int row,int col,int*hash,vector<vector<int> > &res)
/*......*/ {
/*.........*/ int top = getTop(row);
/*.........*/ int left = getLeft(col);
/*.........*/ for (int i = row - 1;i >=0;i--) // 既有结果列中的数不能选
/*.........*/ {
/*...............*/ hash[res[i][col]] = 1;
/*.........*/ }
/*.........*/ if (row > top)
/*.........*/ {
/*............*/ int count = row - top;
/*............*/ for (int i = 0;i <count;i++)
/*...............*/ for (int j = 0;j <3;j++)
/*..................*/hash[res[top+i][left+j]] = 1;
/*.........*/ }
/*......*/ }
/*......*/ voidinit(vector<vector<char> > &board)
/*......*/ {
/*.........*/ int len1 = board.size();
/*.........*/ int len2 = board[0].size();
/*.........*/ for (int i = 0;i <len1;i++)
/*............*/ for (int j = 0;j <len2;j++)
/*............*/ {
/*...............*/ if (board[i][j] != *.*)
/*...............*/ {
/*..................*/ setRow(i,board[i][j]- *0*);
/*..................*/ setCol(j,board[i][j]- *0*);
/*..................*/setCel(i,j,board[i][j] - *0*);
/*..................*/ sel[i][j][0] =board[i][j] - *0*;/*...*/
/*...............*/ }
/*...............*/ else
/*..................*/ sel[i][j][0] = 0;
/*............*/ }
/*......*/ }
/*......*/ void setRow(int i,int num)
/*......*/ {
/*.........*/ for (int j = 0;j < 9;j++)
/*.........*/ {
/*............*/ sel[i][j][num] = 0;
/*.........*/ }
/*......*/ }
/*......*/ void setCol(int j,int num)
/*......*/ {
/*.........*/ for (int i = 0;i < 9;i++)
/*.........*/ {
/*............*/ sel[i][j][num] = 0;
/*.........*/ }
/*......*/ }
/*......*/ void setCel(int x,int y,intnum)// 求方格的左上坐标
/*......*/ {
/*.........*/ int top,left;
/*.........*/ top = getTop(x);
/*.........*/ left = getLeft(y);
/*.........*/ for (int i = 0;i < 3;i++)
/*............*/ for (int j = 0;j <3;j++)
/*...............*/ sel[i+top][j+left][num]= 0;
/*......*/
// 该函数同getTop一样,用来求位于x行y列的格子所在的正方形格的位置。
/*......*/ int getLeft(int col)
/*......*/ {
/*.........*/ int tmp;
/*.........*/ int left;
/*.........*/ tmp = col % 9;
/*.........*/ if (tmp >= 6)
/*.........*/ {
/*............*/ left = 6;
/*.........*/ }
/*.........*/ else if (tmp >= 3)
/*............*/ left = 3;
/*.........*/ else
/*............*/ left = 0;
/*.........*/ return left;
/*......*/ }
/*......*/ int getTop(int row)
/*......*/ {
/*.........*/ int tmp;
/*.........*/ int top;
/*.........*/ tmp = row % 9;
/*.........*/ if (tmp >= 6)
/*.........*/ {
/*............*/ top = 6;
/*.........*/ }
/*.........*/ else if (tmp >= 3)
/*.........*/ {
/*............*/ top = 3;
/*.........*/ }
/*.........*/ else
/*............*/ top = 0;
/*.........*/ return top;
/*......*/ }
};
// 测试函数
int main(void)
{
/*...*/ vector<char> now;
/*...*/ vector<vector<char> >res;
/*...*/ int c;
/*...*/ Solution *p = new Solution();
/*...*/ while ((c = getchar()) != EOF)
/*...*/ {
/*......*/ if (c != ‘\n‘)
/*.........*/ now.push_back(c);
/*......*/ else
/*......*/ {
/*.........*/ res.push_back(now);
/*.........*/ now.clear();
/*......*/ }
}
/*...*/ p->solveSudoku(res);
/*...*/ return 0;
}
【原创】一个基于简单剪枝的DFS解数独程序