Sudoku POJ - 2676

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
DFS。就是需要狂暴剪枝。
一开始只是写了个判断函数判断某个数填在该位置的可行性，然而会T。
因此需要改写成判断在该位置能填哪个数即可。如果一个都不能填则return。
原本冬训的那个数独题由于空格很多。因此还是会T，这需要用到Dancing Links（明天一定学qwq）。
所以就先做了这道……（但我也t了快3小时岂可修）
AC代码如下：

#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
int t,len,cnt=1,vis[10],flag,next[81][2],map[15][15],possible_num[81][10];//记录下一个可以填的位置，0为x轴，1为y轴 
char map1[15][15];
int check(int row,int arr){
    int temp=0;
    memset(vis,0,sizeof(vis));
    for(int i=0;i<9;i++){
        vis[map[row][i]]=1;
        vis[map[i][arr]]=1;
    } 
    for(int i=(row/3)*3;i<(row/3)*3+3;i++)
        for(int j=(arr/3)*3;j<(arr/3)*3+3;j++) 
                vis[map[i][j]]=1;
    for(int i=1;i<=9;i++)
        if(!vis[i]){
            temp++;
            possible_num[cnt][temp]=i;
        }
    return temp;
}
void dfs(int x,int y){
    if(flag)
        return;
    int temp=check(x,y);
    if(temp)
        for(int i=1;i<=temp;i++){
            map[x][y]+=possible_num[cnt][i];
            cnt++;
            if(cnt==len+1)
                flag=1;
            else
                dfs(next[cnt-1][0],next[cnt-1][1]);
            if(flag)
                return;
            cnt--;
            map[x][y]=0;
        }
}
int main(){
    scanf("%d",&t);
    while(t--){
        cnt=1,len=0,flag=0;
        memset(next,0,sizeof(next));
        for(int i=0;i<9;i++)
            for(int j=0;j<9;j++){
                cin>>map1[i][j];
                map[i][j]=map1[i][j]-'0';
                if(!map[i][j]){
                    next[len][0]=i;
                    next[len][1]=j;
                    len++;
                }
            }
        dfs(next[0][0],next[0][1]);
        for(int i=0;i<9;i++){
            for(int j=0;j<9;j++)
                printf("%d",map[i][j]);    
            printf("\n");
        }
    }
    return 0;
}

 

标签：Sudoku,int,cells,2676,len,next,POJ,each,include
来源： https://www.cnblogs.com/mikku39/p/14449108.html