POJ 2676-Sudoku（DFS）

标签：搜索

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

Output

Sample Input

1
103000509
002109400
000704000
300502006
060000050
700803004
000401000
009205800
804000107

Sample Output

143628579
572139468
986754231
391542786
468917352
725863914
237481695
619275843
854396127

深搜。。

题意：就是传统的数独游戏。。找出一种方案即可。。有spj

对 行，列，以及9个3*3的方块哈希。。然后暴力深搜就行了。。实践证明倒搜比较快

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <string>
#include <cctype>
#include <vector>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#define ll long long
#define maxn 360
#define pp pair<int,int>
#define INF 0x3f3f3f3f
#define max(x,y) ( ((x) > (y)) ? (x) : (y) )
#define min(x,y) ( ((x) > (y)) ? (y) : (x) )
using namespace std;
char ma[11][11];
bool row[11][11],col[11][11],mar[11][11],ok;
int change(int x,int y)
{
	if(x>=1&&x<=3) return y%3==0?y/3:y/3+1;
	else if(x>=4&&x<=6) return 3+(y%3==0?y/3:y/3+1);
	else return 6+(y%3==0?y/3:y/3+1);
}
void dfs(int num)
{
	if(ok) return ;
	if(num==0)
	{
		for(int i=1;i<=9;i++)
		{
			for(int j=1;j<=9;j++)
				putchar(ma[i][j]);
			puts("");
		}
		ok=1;
		return ;
	}
	int x=num%9==0?num/9:num/9+1;
	int y=num%9==0?9:num%9;
	if(ma[x][y]-'0')
		dfs(num-1);
	else
	{
		for(int i=1;i<=9;i++)
		{
			if(!row[x][i]&&!col[y][i]&&!mar[change(x,y)][i])
			{
				ma[x][y]=i+'0';row[x][i]=1;col[y][i]=1;mar[change(x,y)][i]=1;
				dfs(num-1);
				ma[x][y]='0';row[x][i]=0;col[y][i]=0;mar[change(x,y)][i]=0;
			}
		}
	}
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		memset(row,0,sizeof(row));
		memset(col,0,sizeof(col));
		memset(mar,0,sizeof(mar));
		for(int i=1;i<=9;i++)
			scanf("%s",ma[i]+1);
			for(int i=1;i<=9;i++)
				for(int j=1;j<=9;j++)
				if(ma[i][j]-'0')
				{
					row[i][ma[i][j]-'0']=1;
					col[j][ma[i][j]-'0']=1;
					mar[change(i,j)][ma[i][j]-'0']=1;
				}
		ok=0;dfs(81);
	}
	return 0;
}

POJ 2676-Sudoku（DFS）

标签：搜索

原文地址：http://blog.csdn.net/qq_16255321/article/details/41366269