Sudoku Killer

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1293 Accepted Submission(s): 424

Problem Description Since the first several Sino-World Championships from March 10 to 11, 2006, the number of this game is increasingly loved and paid attention to. It is said that at the 2008 Beijing Olympics, the number is alone as a separate project, and the champion will have a huge prize --- HDU free seven-day tour Extra Lcy checking and chance to take a photo with HDU ACM Team. So before the world, the servant is successively trained for the prizes day and night. Of course, it also includes the beginner Linle, but he is too stupid and there is not much patience, can only do the most basic number of autism, but he still wants to get those prizes, can you help him? As long as you tell him about the answer, you don't have to teach him how to do it. The rule of the number of games is this: in a 9x9 square, you need to fill in the number 1-9 to the space, and make the checkered each line and each column. They all contain 1-9 nine numbers. At the same time, it is also necessary to divide the thick line in the space into 9 3x3 squares, also contain 1-9 nine numbers. For example, there is such a question, you can take a closer look at each line, each column, and each 3X3 checker contains 1-9 nine numbers. Example: answer:

Input This topic contains multiple sets of tests, separated from one space between each group. Each group of tests will give you a 9 * 9 matrix, and the two elements adjacent to the same line are separated by a space. Among them, 1-9 represents the number of filledments in this location, question mark (?) Means you need to fill the number.

Output For each group test, then output its solution, and the two numbers adjacent to the same line are separated by a space. There is an empty line between the two groups. For each set of test data, it is guaranteed to have only one solution.

Sample Input 7 1 2 ? 6 ? 3 5 8 ? 6 5 2 ? 7 1 ? 4 ? ? 8 5 1 3 6 7 2 9 2 4 ? 5 6 ? 3 7 5 ? 6 ? ? ? 2 4 1 1 ? 3 7 2 ? 9 ? 5 ? ? 1 9 7 5 4 8 6 6 ? 7 8 3 ? 5 1 9 8 5 9 ? 4 ? ? 2 3

Sample Output 7 1 2 4 6 9 3 5 8 3 6 5 2 8 7 1 9 4 4 9 8 5 1 3 6 7 2 9 2 4 1 5 6 8 3 7 5 7 6 3 9 8 2 4 1 1 8 3 7 2 4 9 6 5 2 3 1 9 7 5 4 8 6 6 4 7 8 3 2 5 1 9 8 5 9 6 4 1 7 2 3

Author linle

Source ACM Summer Training Team Exercise (3)

Recommend LL

This question is to input and output, it is easy to go wrong.

I just started to use% C, every time I have to get Char () to go to space and enter;

Later, I used to use% s with% s.

This output is the output space between the two groups, and the last set of data does not have to output.

Because there is no direct return after finding, the results are not right, and I will find it wrong for a long time.

#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<algorithm>  
#include<cmath>  
using namespace std;
char map[10][10];

int x[100];
int y[100];

int vx[10][10];
int vy[10][10];
int vz[10][10];

int cnt;
bool flag;

void dfs(int deep)
{
	if (flag)return;
	if (deep == cnt) { flag = true; return; }
		int i = deep;
		for (int j = 1; j <= 9; j++)
		{
			IF (flag) return; // less
			int temp = 0;
			 IF (vx [x [i]] [j] == 1) Continue; // is not full
			 IF (Vy [Y]] [J] == 1) Continue; // Not satisfied
			 IF (x [i]> = 0 && x [i] <3) // does not satisfy 3 * 3
			{
				if (y[i] >= 0 && y[i] < 3)
				{
					temp = 0;
					if(vz[0][j] == 1)continue;
				}
				else	if (y[i] >= 3 && y[i] < 6)
				{
					temp = 1;
					if (vz[1][j] == 1)continue;
				}
				else	if (y[i] >= 6 && y[i]< 9)
				{
					temp = 2;
					if (vz[2][j] == 1)continue;
				}
			}
			else	if (x[i] >= 3 && x[i] < 6)
			{
				if (y[i] >= 0 && y[i] < 3)
				{
					temp = 3;
					if (vz[3][j] == 1)continue;
				}
				else	if (y[i] >= 3 && y[i] < 6)
				{
					temp = 4;
					if (vz[4][j] == 1)continue;
				}
				else	if (y[i] >= 6 && y[i]< 9)
				{
					temp = 5;
					if (vz[5][j] == 1)continue;
				}
			}
			else	if (x[i] >= 6 && x[i] < 9)
			{
				if (y[i] >= 0 && y[i]< 3)
				{
					temp = 6;
					if (vz[6][j] == 1)continue;
				}
				else	if (y[i] >= 3 && y[i] < 6)
				{
					temp = 7;
					if (vz[7][j] == 1)continue;
				}
				else	if (y[i] >= 6 && y[i]< 9)
				{
					temp = 8;
					if (vz[8][j] == 1)continue;
				}
			}
			map[x[i]][y[i]] = j+'0';
			vx[x[i]][j] = 1;
			vy[y[i]][j] = 1;
			vz[temp][j] = 1;
			dfs(deep + 1);
			vx[x[i]][j] = 0;
			vy[y[i]][j] = 0;
			vz[temp][j] = 0;
		}
	

}


int main()
{
	int l = 0;
	char str[10];
	while (scanf("%s", &str) != EOF)
	{
		map[0][0] = str[0];
	
		for (int i = 0; i < 9; i++)
		{
			for (int j = 0; j < 9; j++)
			{
				if (i == 0 && j == 0)continue;
				scanf("%s", &str);
				map[i][j] = str[0];
			}
			
		}
		cnt = 0;
		memset(vx, 0, sizeof(vx));
		memset(vy, 0, sizeof(vy));
		memset(vz, 0, sizeof(vz));
		for (int i = 0; i < 9; i++)
		{
			for (int j = 0; j < 9; j++)
			{
				if (map[i][j] == '?') { x[cnt] = i; y[cnt++] = j; }
				else 
				{
					int temp = map[i][j] - '0';
					vx[i][temp] = 1;
					vy[j][temp] = 1;
					if (i >= 0 && i < 3)
					{
						if (j >= 0 && j < 3)
						{
							vz[0][temp] = 1;
						}
						else	if (j >= 3 && j < 6)
						{
							vz[1][temp] = 1;
						}
						else	if (j >= 6 &&  j< 9)
						{
							vz[2][temp] = 1;
						}
					}
					else	if (i >= 3 && i < 6)
					{
						if (j >= 0 && j < 3)
						{
							vz[3][temp] = 1;
						}
						else	if (j >= 3 && j < 6)
						{
							vz[4][temp] = 1;
						}
						else	if (j >= 6 && j< 9)
						{
							vz[5][temp] = 1;
						}
					}
					else	if (i >= 6 && i < 9)
					{
						if (j >= 0 && j < 3)
						{
							vz[6][temp] = 1;
						}
						else	if (j >= 3 && j < 6)
						{
							vz[7][temp] = 1;
						}
						else	if (j >= 6 && j< 9)
						{
							vz[8][temp] = 1;
						}
					}
				}
			}
		}
		flag = false;
		if (l)printf("\n");
		dfs(0);
		l++;
		for (int i = 0; i < 9; i++)
		{
			for (int j = 0; j < 8; j++)
			{
				printf("%c ", map[i][j]);
			}
			printf("%c\n", map[i][8]);

		}

	}
	return 0;
}