----------------------------------------------------------------------------
Mean:
求解數獨.
analyse:
隻是9宮格的數獨,而且測試資料都不難,是以可以直接使用遞歸求解,類似于N-Queue問題.
但如果宮格數較多,則需要使用Dancing-Link精确覆寫算法來求解.
Time complexity: O(N)
view code
/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* Author: crazyacking
* Date : 2016-03-02-18.53
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <bits/stdc++.h>
#include <windows.h>
using namespace std;
typedef long long(LL);
typedef unsigned long long(ULL);
const double eps(1e-8);
class Solution
{
public:
void solveSudoku(vector<vector<char>>& board)
{
recursiveSolve(board);
}
bool recursiveSolve(vector<vector<char>>& board)
for(int i=0;i<9;++i)
{
for(int j=0;j<9;++j)
{
if(board[i][j]=='.')
{
for(int k=1;k<=9;++k)
{
board[i][j]=static_cast<char>(k+'0');
if(isValid(board,i,j) && recursiveSolve(board))
return true;
board[i][j]='.';
}
return false;
}
}
}
return true;
bool isValid(const vector<vector<char>>& board,const int r1,const int c1)const
if(i!=r1 && board[i][c1]==board[r1][c1])
return false;
if(i!=c1 && board[r1][i]==board[r1][c1])
int rowBegin=r1/3*3;
int colBegin=c1/3*3;
for(int i=rowBegin;i<rowBegin+3;++i)
for(int j=colBegin;j<colBegin+3;++j)
if(i!=r1 && j!=c1 && board[i][j]==board[r1][c1])
};
int main()
freopen("H:\\Code_Fantasy\\in.txt","r",stdin);
Solution solution;
vector<vector<char>> ve;
string s;
while(cin>>s)
vector<char> tempVe;
for(int i=0;i<s.length();++i)
tempVe.push_back(s[i]);
ve.push_back(tempVe);
solution.solveSudoku(ve);
return 0;
}
/*