leetcode-Sudoku Solver

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.

题意：给出一个数独，求解，保证有唯一解

分析：简单的思路是，遍历每个空格，枚举所有可能取值，递归求解。为了减小搜索空间，用启发式搜索。即，记录每个空格可能取值的种类，按从小到大遍历所有空格， 保证搜索空间最小。用三个set[9]分别记录每行，每列，每个3*3区域已有的数字，然后就能在O(1)内对相应的set做并集，求出任一空格可能的取值。

代码：

int cmp(const void *a, const void *b)
{
    return ((int*)a)[2]-((int*)b)[2];
}

class Solution {
private:
    set<char> row[9],col[9],batch[9];
    int op_qu[90][3],qu_cnt;


public:
    bool fill(vector<vector<char> > &bd,int num)
    {
        if(num>=qu_cnt) return true;
        int i=op_qu[num][0],j=op_qu[num][1];
        set<char> tmp,now;
        set_union(row[i].begin(),row[i].end(),col[j].begin(),col[j].end(),inserter(tmp,tmp.begin()));
        int k = i/3*3+j/3;
        set_union(tmp.begin(),tmp.end(),batch[k].begin(),batch[k].end(),inserter(now,now.begin()));

        for(char ch='1'; ch<='9'; ch++)
        {
            if(now.find(ch)==now.end())
            {
                bd[i][j] = ch;
                row[i].insert(ch);
                col[j].insert(ch);
                batch[k].insert(ch);
                if(fill(bd,num+1))
                    return true;
                row[i].erase(ch);
                col[j].erase(ch);
                batch[k].erase(ch);
            }
        }
        return false;
    }
    void solveSudoku(vector<vector<char> > &board) {

        for(int i=0; i<9; i++)
        {
            row[i].clear();
            col[i].clear();
            batch[i].clear();
        }
        qu_cnt = 0;

        for(int i=0; i<9; i++)
        {
            for(int j=0; j<9; j++)
            {
                char ch = board[i][j];
                if(ch!='.')
                {
                    row[i].insert(ch);
                    col[j].insert(ch);
                    batch[i/3*3+j/3].insert(ch);
                }
            }
        }
        for(int i=0; i<9; i++)
        {
            for(int j=0; j<9; j++)
            {
                char ch = board[i][j];
                if(ch=='.')
                {
                    set<char> tmp,now;
                    set_union(row[i].begin(),row[i].end(),col[j].begin(),col[j].end(),inserter(tmp,tmp.begin()));
                    int k = i/3*3+j/3;
                    set_union(tmp.begin(),tmp.end(),batch[k].begin(),batch[k].end(),inserter(now,now.begin()));

                    op_qu[qu_cnt][0] = i;
                    op_qu[qu_cnt][1] = j;
                    op_qu[qu_cnt][2] = 9-now.size();
                    qu_cnt++;
                }
            }
        }
        qsort(op_qu,qu_cnt,sizeof(int)*3,cmp);

        fill(board,0);
    }
};