POJ 2676 数独 搜索

Sudoku

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 16386 Accepted: 8004 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

1103000509002109400000704000300502006060000050700803004000401000009205800804000107

Sample Output

143628579572139468986754231391542786468917352725863914237481695619275843854396127

Source

Southeastern Europe 2005

题意：给9个3*3的小正方形，使得小正方形里恰好有9个不重复的1--9，而且大正方形行或列不能有重复的数字。

题解：不太容易的搜索，开始根本没思路，还是看了课件的代码理解了一遍才开始敲得。说一下思路吧。开二维数组col，row分别记录大正方形所有的行，列数字出现情况，block记录9个小正方形的数字出现情况，主要是判断重复，加快搜索速度。这里给出的小矩阵的计算公式为(r/3)*3+c/3，可以表示为第几个矩阵。

这里我们同样要用到递归的思想，先标记为访问过，结束递归再恢复状态，所有的点都被赋予值则可以判断矩阵构造好了。

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<algorithm>using namespace std;#define CLR(x) memset((x),0,sizeof(x));#define N 20char ss[N][N];int g[N][N],col[N][N],row[N][N],block[N][N];struct point{int c,r;point (int _r,int _c):r(_r),c(_c){}point(){}};vector<point>eg;inline int get_block(int r,int c){r/=3;c/=3;return (r*3+c);}inline void SET(int r,int c,int num,int n){col[c][num]=block[get_block(r,c)][num]=row[r][num]=n;}inline bool judge(int r,int c,int num){return(!col[c][num])&&(!row[r][num])&&(!block[get_block(r,c)][num]);}bool dfs(int n){if(n<0){return true;}int r=eg[n].r;int c=eg[n].c;for(int i=1;i<=9;i++){if(judge(r,c,i)){g[r][c]=i;SET(r,c,i,1);if(dfs(n-1))return true;//在这里构造成功会不断跳出递归SET(r,c,i,0);}}return false;}int main(){#ifdef CDZSCfreopen("i.txt","r",stdin);#endifint t;scanf("%d",&t);while(t--){eg.clear();CLR(g);CLR(col);CLR(row);CLR(block);for(int i=0;i<9;i++){scanf("%s",ss[i]);}for(int i=0;i<9;i++){for(int j=0;j<9;j++){if(ss[i][j]-'0'==0){eg.push_back(point(i,j));}else{g[i][j]=ss[i][j]-'0';SET(i,j,ss[i][j]-'0',1);}}}if(dfs(eg.size()-1)){for(int i=0;i<9;i++){for(int j=0;j<9;j++){printf("%d",g[i][j]);}printf("\n");}}}return 0;}