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);
}
};