数独-poj2676-dfs(神（深）搜)

Sudoku

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 21955		Accepted: 10412		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

1
103000509
002109400
000704000
300502006
060000050
700803004
000401000
009205800
804000107

Sample Output

143628579
572139468
986754231
391542786
468917352
725863914
237481695
619275843
854396127

Source

Southeastern Europe 2005

题意：数独填数字，每一行，每一列，每一块1到9数字不能重复，求解。dfs,，把每个需要填的位置记录下来，dfs结束条件就为填的位置都填完了就输出，注意返回条件很重要，原来落一个返回，结果没有输出了：）

代码：

#include<stdio.h>
#include<vector>
#include<string.h>
using namespace std;
bool row[10][10];
bool cal[10][10];
bool  kuai[10][10];
int a[10][10],n;
struct dian
{
    int i,j;
}d;
vector<dian>f;
bool flag;
void dfs(int k)
{

    if(k==n)
    {
        flag=true;
        for(int o=0;o<9;o++)
        {
            for(int u=0;u<9;u++)
            {
                printf("%d",a[o][u]);
            }
            printf("\n");
        }
        return;
    }
    for(int i=1;i<=9;i++)
    {
        dian g=f[k];
        int p=(g.i/3)*3+g.j/3+1;//找到对应位置是第几块，至于为什么画个图算一下很容易看出来
        if(row[g.i][i]==0&&cal[g.j][i]==0&&kuai[p][i]==0)
        {
            row[g.i][i]=cal[g.j][i]=kuai[p][i]=1;
            a[g.i][g.j]=i;
            dfs(k+1);
            if(flag)return;//最关键是这里，如果不及时返回则找到的答案就会前功尽弃
            row[g.i][i]=cal[g.j][i]=kuai[p][i]=0;
            a[g.i][g.j]=0;
        }
    }
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
    f.clear();
    memset(cal,0,sizeof(cal));
    memset(row,0,sizeof(row));
    memset(kuai,0,sizeof(kuai));
    for(int i=0;i<9;i++)
    {
        for(int j=0;j<9;j++)
        {
            scanf("%1d",&a[i][j]);
            if(a[i][j])
            {
                row[i][a[i][j]]=1;
                cal[j][a[i][j]]=1;
                int k=(i/3)*3+j/3+1;
                kuai[k][a[i][j]]=1;
            }
            else
            {
                d.i=i;
                d.j=j;
                f.push_back(d);
            }
        }
    }
    flag=false;
     n=f.size();
    dfs(0);//这里注意要从0开始，因为vector里是从0开始的，否则就会少点


}
}