poj 3074 Sudoku

In the game of Sudoku, you are given a large 9 × 9 grid divided into smaller 3 × 3 subgrids. For example,

Given some of the numbers in the grid, your goal is to determine the remaining numbers such that the numbers 1 through 9 appear exactly once in (1) each of nine 3 × 3 subgrids, (2) each of the nine rows, and (3) each of the nine columns.

The input test file will contain multiple cases. Each test case consists of a single line containing 81 characters, which represent the 81 squares of the Sudoku grid, given one row at a time. Each character is either a digit (from 1 to 9) or a period (used to indicate an unfilled square). You may assume that each puzzle in the input will have exactly one solution. The end-of-file is denoted by a single line containing the word “end”.

For each test case, print a line representing the completed Sudoku puzzle.

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
const int tmp[9][9] = {{0,0,0,1,1,1,2,2,2},
					   {0,0,0,1,1,1,2,2,2},
					   {0,0,0,1,1,1,2,2,2},
					   {3,3,3,4,4,4,5,5,5},
					   {3,3,3,4,4,4,5,5,5},
					   {3,3,3,4,4,4,5,5,5},
					   {6,6,6,7,7,7,8,8,8},
					   {6,6,6,7,7,7,8,8,8},
				 	   {6,6,6,7,7,7,8,8,8}};
int Int(){
    char ch;int tmp = 0;
    while (ch = getchar()) if (ch >= '0' && ch <='9') break;
    for (; ch >= '0' && ch <= '9'; ch = getchar()) tmp = tmp * 10 + ch - '0';
	return tmp;
}
int tmp1[9], tmp2[9], tmp3[9];
int a[9][9];
int ans[9][9];
bool ok;
int num[1 << 12];
void init(){
	memset(tmp1, 0, sizeof(tmp1));
	memset(tmp2, 0, sizeof(tmp2));
	memset(tmp3, 0, sizeof(tmp3));
	memset(ans, 0, sizeof(ans));
	memset(a, 0, sizeof(a));
	ok = false;
	for (int i = 0; i < 9; i ++)
		for (int j = 0; j < 9; j ++){
			char ch = 0;
			while (ch = getchar()){
				if ((ch >= '0' && ch <= '9') || ch == '.') break;
				if (ch == 'e') exit(0);
			}
			if (ch != '.') a[i][j] = ch - '0';
			if (a[i][j]){
				ans[i][j] = a[i][j];
				tmp1[i] ^= (1 << a[i][j] - 1);
				tmp2[j] ^= (1 << a[i][j] - 1);
				tmp3[tmp[i][j]] ^= (1 << a[i][j] - 1);
			}
		}
}
int find_next(int &x, int &y){
	x = 9, y = 0;
	int mini = 99999;
	for (int i = 0; i < 9; i ++)
		for (int j = 0; j < 9; j ++)
			if (ans[i][j] == 0 && num[(511 ^ tmp1[i]) & (511 ^ tmp2[j]) & (511 ^ tmp3[tmp[i][j]])] < mini){
				x = i, y = j;
				mini = num[(511 ^ tmp1[i]) & (511 ^ tmp2[j]) & (511 ^ tmp3[tmp[i][j]])];
			}
}
void dfs(int x, int y){
	if (ok)return;
	if (a[x][y]){
		int xx = 0, yy = 0;
		find_next(xx, yy); 
		dfs(xx, yy);
		return;
	}
	if (x == 9 && y == 0){
		for (int i = 0; i < 9; i ++){
			for (int j = 0; j < 9; j ++)
				cout <<ans[i][j] ;	
		}
		cout <<endl;
		ok = true;
		return;
	}
	for (int i = 1; i <= 9; i ++)
		if (!(tmp1[x] & (1 << i - 1)) && !(tmp2[y] & (1 << i - 1)) && !(tmp3[tmp[x][y]] & (1 << i - 1))){
			ans[x][y] = i;
			tmp1[x] ^= (1 << i - 1);
			tmp2[y] ^= (1 << i - 1);
			tmp3[tmp[x][y]] ^= (1 << i - 1);
			int xx = 0, yy = 0;
			find_next(xx, yy); 
			dfs(xx, yy);
			ans[x][y] = 0;
			tmp1[x] ^= (1 << i - 1);
			tmp2[y] ^= (1 << i - 1);
			tmp3[tmp[x][y]] ^= (1 << i - 1);
		}	
}
int main(){
	for (int i = 0; i <= 512; i ++)
		for (int j = 0; j < 9; j ++)
			if (i & (1 << j)) num[i] ++;
	while (1){ 
		init();
		dfs(0, 0);
	}
	return 0;
}