poj2676 sudoku

最新推荐文章于 2022-08-11 20:58:55 发布

转载最新推荐文章于 2022-08-11 20:58:55 发布 · 49 阅读

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原文链接：http://www.cnblogs.com/13224ACMer/p/4764593.html

本文介绍了一种使用深度优先搜索解决数独问题的算法。通过记录每行、每列及每个九宫格内的数字出现情况，确保填入的每一个数字都符合数独规则。采用递归方式填充空缺单元格，最终输出解决方案。

Sudoku

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 16102		Accepted: 7867		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

Source

Southeastern Europe 2005

数独，提议就不用解释了吧。。。。

我用row[k][i]表示第k行是否已经有数i

col[k][i]表示第k列是否已经有数i

而sq[k][i]表示第k个九宫格是否有数i（这里k=(x/3)*3+(y/3),可以想想为什么）

从0,0 依次搜到8,8 DFS依次往下，看代码：（已有注释）

#include<stdio.h>

#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int map[10][10];
bool hang[10][10],lie[10][10],sqr[10][10];

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


bool flag;
void bfs(int x,int y){
   int u=x*9+y+1;
   if(x==9){
       flag=true;
       print();
       return ;
   }
   if(flag)
   return ;
   if(map[x][y]!=-1){
       bfs(u/9,u%9);
       return ;
   }
   for(int i=0;i<9&&!flag;i++){
       if(!hang[x][i]&&!lie[y][i]&&!sqr[x/3*3+y/3][i]){
           hang[x][i]=true;
           lie[y][i]=true;
           sqr[x/3*3+y/3][i]=true;
           map[x][y]=i;
           bfs(u/9,u%9);
           hang[x][i]=false;
           lie[y][i]=false;
           map[x][y]=-1;
           sqr[x/3*3+y/3][i]=false;
       }
   }

}


int main(){
  int t;
  scanf("%d",&t);
  while(t--){
     char c;
     memset(hang,false,sizeof(hang));
     memset(lie,false,sizeof(lie));

     memset(map,0,sizeof(map));
     memset(sqr,false,sizeof(sqr));
     flag=false;
     for(int i=0;i<9;i++){
     for(int j=0;j<9;j++){
        cin>>c;
        map[i][j]=c-'1';
        if(c>='1'&&c<='9'){
            int temp=c-'1';
            hang[i][temp]=true;
            lie[j][temp]=true;

            sqr[i/3*3+j/3][temp]=true;
        }
     }
    }
    bfs(0,0);
  }
  return 0;
}

转载于:https://www.cnblogs.com/13224ACMer/p/4764593.html