sicily 1162. Sudoku

最新推荐文章于 2015-12-30 18:02:00 发布

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本文介绍了一种使用深度搜索算法解决Sudoku谜题的方法，详细解释了算法的实现过程，包括如何标记已有的数字以避免重复，并通过三个标记数组来检查每一行、每一列和每一个3x3的小宫格内是否已经包含特定的数字。同时，文章确保了算法的正确性和效率，确保了在有限的时间和内存限制下能够找到解。

1162. Sudoku

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB , Special Judge

Description

Sudoku is a very simple task. A square table with 9 rows and 9 columns is divided to 9 smaller squares 3x3 as shown on the Figure. In some of the cells are written decimal digits from 1 to 9. The other cells are empty. The goal is to fill the empty cells with decimal digits from 1 to 9, one digit per cell, in such way that in each row, in each column and in each marked 3x3 subsquare, all the digits from 1 to 9 to appear. Write a program to solve a given Sudoku-task.

Input

The input data will start with the number of the test cases. For each test case, 9 lines follow, corresponding to the rows of the table. On each line a string of exactly 9 decimal digits is given, corresponding to the cells in this line. If a cell is empty it is represented by 0.

Output

For each test case your program should print the solution in the same format as the input data. The empty cells have to be filled according to the rules. If solutions is not unique, then the program may print any one of them.

Sample Input

Sample Output

题目分析

看题目,深搜

注意数字要在每行,每列,每个块中唯一

直接深搜注意两点,否则超时

一,标记为零的区域,而不是整个盘扫一遍去判断某个位置是否需要填数

二,开三个标记数组,确定在某行,某列,某小宫格是否已经有了某个数字

题目确保有解,所以一旦填完所有的空即可返回

#include <cstdio>
#include <iostream>
#include <memory.h>

bool inRow[10][10]; // if inRow[i][j] == true, means there is a digit --j, in ith row
bool inCol[10][10];
bool inBlock[10][10];
int suduku[10][10];
int test[81][2]; // record the position of the undefined cells
int count; // record the num of the undefined cells
bool doit;

int getBlockId(int x, int y) {
  return (x-1)/3*3 + (y-1)/3 + 1;
}

void dfs(int index) {
  int row = test[index][0];
  int col = test[index][1];
  for (int i = 1; i <= 9 && !doit; ++i) {
    if (inRow[row][i] || inCol[col][i] || inBlock[getBlockId(row,col)][i])
      continue;
    suduku[row][col] = i;
    inRow[row][i] = true;
    inCol[col][i] = true;
    inBlock[getBlockId(row,col)][i] = true;
    ++index;
    if (index == count) {
      doit = true;
      return;
    }
    dfs(index);
    if (doit) {
      return;
    } else {
      --index;
      suduku[row][col] = 0;
      inRow[row][i] = false;
      inCol[col][i] = false;
      inBlock[getBlockId(row,col)][i] = false;
    }
  }
}

int main() {
  int num;
  scanf("%d", &num);
  while (num--) {
    memset(inRow, 0, sizeof(inRow));
    memset(inCol, 0, sizeof(inCol));
    memset(inBlock, 0, sizeof(inBlock));
    count = 0;
    doit = false;
    std::string str;
    for (int i = 1; i <= 9; ++i) {
      std::cin >> str;
      for (int j = 1; j <= 9; ++j) {
        suduku[i][j] = str[j-1] - '0';
        if (suduku[i][j]) {
          inRow[i][suduku[i][j]] = true;
          inCol[j][suduku[i][j]] = true;
          inBlock[getBlockId(i,j)][suduku[i][j]] = true;
        } else {
          test[count][0] = i;
          test[count][1] = j;
          count++;
        }
      }
    }

    dfs(0);
    for (int i = 1; i <= 9; ++i) {
      for (int j = 1; j <= 9; ++j)
        printf("%d", suduku[i][j]);
      printf("\n");
    }
  }
}