Leetcode 37 Sudoku Solver 深搜基础题+位运算

Write a program to solve a Sudoku puzzle by filling the empty cells.

Empty cells are indicated by the character '.'.

You may assume that there will be only one unique solution.

A sudoku puzzle...

...and its solution numbers marked in red.

强烈推荐的算法入门题，填数独，深搜方法完成，采用的标记方法依然和上一题一样，位操作。

http://blog.csdn.net/accepthjp/article/details/52448027

注意回溯

class Solution {public:    bool dfs(int dep,vector<vector<char>>& board,int vis[],int vis2[],int vis3[])    {        if(dep==81) return true; //全部填完        int x=dep/9,y=dep%9;        if(board[x][y]!='.')            return dfs(dep+1,board,vis,vis2,vis3);        else        {            for(int i=1;i<10;i++)            {                if(vis[x]&(1<<i) || vis2[y]&(1<<i) || vis3[x/3*3+y/3]&(1<<i)) continue;//不能填写该数字                vis[x]|=(1<<i);                vis2[y]|=(1<<i);                vis3[x/3*3+y/3]|=(1<<i);                if(dfs(dep+1,board,vis,vis2,vis3)) //可以填写该数字                {                    board[x][y]='0'+i;                    return true;                }                vis[x]&=~(1<<i);                vis2[y]&=~(1<<i);                vis3[x/3*3+y/3]&=~(1<<i);            }        }        return false; //此位置无数可填    }    void solveSudoku(vector<vector<char>>& board)     {        int vis[9]={0},vis2[9]={0},vis3[9]={0};        for(int i=0;i<9;i++)                     //预处理已经填好的位置            for(int j=0;j<9;j++)                if(board[i][j]!='.')                {                    int t=board[i][j]-'0';                    vis[i]|=(1<<t);                    vis2[j]|=(1<<t);                    vis3[i/3*3+j/3]=(1<<t);                }        dfs(0,board,vis,vis2,vis3);    }};