poj2676--Sudoku(搜索练习5-数独游戏)
来源:互联网 发布:沥青路面厚度设计软件 编辑:程序博客网 时间:2024/06/06 03:58
Sudoku
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64uAppoint description:
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
找出每一行,每一列,每一个小方格中可以填的数,进行dfs搜索结果
#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;int l[10][10] , r[10][10] , c[10][10] ;struct node{ int x , y ;} p[100] ;int num ;char str[12][12] ;int dfs(int k){ if( k == num ) return 1 ; int x = p[k].x , y = p[k].y ; int i ; for(i = 1 ; i <= 9 ; i++) { if( !r[x][i] && !l[y][i] && !c[x/3*3+y/3][i] ) { r[x][i] = l[y][i] = 1 ; c[x/3*3+y/3][i] = 1 ; str[x][y] = i + '0' ; if( dfs(k+1) ) return 1 ; str[x][y] = '0' ; r[x][i] = l[y][i] = 0 ; c[x/3*3+y/3][i] = 0 ; } } return 0 ;}int main(){ int t , i , j ; scanf("%d", &t) ; while( t-- ) { num = 0 ; memset(l,0,sizeof(l)) ; memset(r,0,sizeof(r)) ; memset(c,0,sizeof(c)) ; for(i = 0 ; i < 9 ; i++) { scanf("%s", str[i]) ; for(j = 0 ; j < 9 ; j++) { if( str[i][j] != '0' ) { r[i][ str[i][j] - '0' ] = 1 ; l[j][ str[i][j] - '0' ] = 1 ; c[ i/3*3+j/3 ][ str[i][j] - '0' ] = 1 ; } else { p[num].x = i ; p[num++].y = j ; } } } dfs(0) ; for(i = 0 ; i < 10 ; i++) printf("%s\n", str[i]) ; } return 0;}
0 0
- poj2676--Sudoku(搜索练习5-数独游戏)
- poj2676(数独 sudoku)
- POJ2676-Sudoku(数独)
- poj2676 Sudoku 数独
- POJ2676 Sudoku [数独]
- poj2676 Sudoku 数独
- (POJ2676)Sudoku <简单数独问题>
- POJ2676--Sudoku(搜索)
- 小算法练习,Sudoku(POJ2676数字游戏
- poj2676数独(dfs)
- POJ2676数独(DFS)
- POJ 2676 Sudoku (数独 搜索)
- 华为OJ(数独游戏-Sudoku)
- LeetCode:Valid Sudoku,Sudoku Solver(数独游戏)
- poj2676——Sudoku(深度搜索)
- POJ2676-搜索-Sudoku
- dfs数独--poj2676
- 简单搜索(数独)poj 2676 Sudoku
- Epoll模型详解
- 利用V4L2编写的USB摄像头程序1【经OK6410验证成功】
- 【cocos2d-js官方文档】十七、事件分发机制
- 在Qt下配置opencv2.4.9
- malloc alignment
- poj2676--Sudoku(搜索练习5-数独游戏)
- DRP之synchronized及锁的思考
- iOS内存管理(一)
- USACO 1.3 Prime Cryptarithm
- dwr函数调用一点小问题
- 五种开源协议的比较(BSD,Apache,GPL,LGPL,MIT) – 整理
- 【Android入门 三】创建项目时,有关appcompat_v7工程报错问题的分析和排除
- 一个衡中走出的小程序员
- Linux下rz/sz安装及使用方法