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 804000107Sample Output
143628579 572139468 986754231 391542786 468917352 725863914 237481695 619275843 854396127思路：dfs+简单剪枝；简单剪枝：只填空白格子，所填数字应该是所在行，列，方块中都没有填过的数字；
代码：
#include <bits/stdc++.h> using namespace std; pair<int,int> p[100]; bool row[10][10], col[10][10], g[4][4][10]; int G[10][10], cnt; bool dfs(int cn) { if(cn < 0) return true; int x = p[cn].first, y = p[cn].second; for(int k=1; k<=9; k++) { if(row[x][k] || col[y][k] || g[x/3][y/3][k])//剪枝：若填过 ，则继续 continue; row[x][k] = col[y][k] = g[x/3][y/3][k] = true; G[x][y] = k; if(dfs(cn-1)) return true; row[x][k] = col[y][k] = g[x/3][y/3][k] = false;//以便计算下一个方案 } return false; } int main() { int T; scanf("%d", &T); while(T--) { cnt = 0; for(int i=0; i<9; i++) { for(int j=0; j<9; j++) { scanf(" ", &G[i][j]); int k = G[i][j]; if(k != 0) { row[i][k] = col[j][k] = g[i/3][j/3][k] = true; } else {//剪枝：保存空白格子坐标 p[cnt].first=i; p[cnt].second=j; cnt++; } } } dfs(cnt-1);//倒着搜 for(int i=0; i<9; i++) { for(int j=0; j<9; j++) { printf("%d", G[i][j]); } putchar('\n'); } } return 0; }
