poj2676——Sudoku

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

解数独的题目
自己写了一大堆越写越乱，最后还是参考了一下题解，感觉写得比较巧妙，把dfs起点的遍历也放在dfs里，而不是枚举起点再dfs

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
bool row[15][15];
bool col[15][15];
bool ge[15][15];
char map[15][15];
int mp[15][15];
int t;
void init(){
    memset(row,false,sizeof(row));
    memset(col,false,sizeof(col));
    memset(ge,false,sizeof(ge));
    memset(mp,'\0',sizeof(mp));
}
bool dfs(int i,int j){
    //9行已经完成
    if(i==9){
        return true;
    }
    bool flag=false;
    //已有数字，搜索下一个位置
    if(mp[i][j]){
        if(j==8){
            //下一行
            flag=dfs(i+1,0);
        }
        else{
            //往右
            flag=dfs(i,j+1);
        }
        //回溯
        if(flag){
            return true;
        }
        else{
            return false;
        }
    }
    //没有数字
    else{
        int k=(i/3)*3+(j+3)/3;
        for(int t=1;t<=9;t++){
            if(!row[i][t] && !col[j][t] && !ge[k][t]){
                mp[i][j]=t;
                row[i][t]=true;
                col[j][t]=true;
                ge[k][t]=true;
                //判断是否换行
                if(j==8){
                    flag=dfs(i+1,0);
                }
                else{
                    flag=dfs(i,j+1);
                }
                //回溯
                if(!flag){
                    //搜索失败，重置
                    mp[i][j]=0;
                    row[i][t]=false;
                    col[j][t]=false;
                    ge[k][t]=false;
                }
                else{
                    return true;
                }
            }
        }
    }
    return false;
}
int main(void){
    //freopen("data.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    scanf("%d",&t);
    while(t--){
        init();
        for(int i=0;i<9;i++){
            scanf("%s",map[i]);
            for(int j=0;j<9;j++){
                mp[i][j]=map[i][j]-'0';
                if(mp[i][j]){
                    row[i][mp[i][j]]=true;
                    col[j][mp[i][j]]=true;
                    ge[(i/3)*3+(j+3)/3][mp[i][j]]=true;
                }
            }
        }
        dfs(0,0);
        for(int i=0;i<9;i++){
            for(int j=0;j<9;j++){
                printf("%d",mp[i][j]);
            }
            printf("\n");
        }
    }
    return 0;
}