标签：style http color os io ar for div 问题
The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.
Given an integer n, return all distinct solutions to the n-queens puzzle.
Each solution contains a distinct board configuration of the n-queens‘ placement, where ‘Q‘ and ‘.‘ both
indicate a queen and an empty space respectively.
For example,
There exist two distinct solutions to the 4-queens puzzle:
[ [".Q..", // Solution 1 "...Q", "Q...", "..Q."], ["..Q.", // Solution 2 "Q...", "...Q", ".Q.."] ]
题意：返回n皇后问题的所有可能解
思路：dfs
要求给出所有可能解的题，一般采用dfs 暴力枚举
这道题设置了额外的三个bool数组分别表示某一列，
某一主对角线，某一副对角线是否已经有皇后占据着。
由某个坐标(i,j)，可得到它的主对角线是i+j，它的副对角线是i-j，
由于i-j可能是负数，所以我们给它加上 n,保证它不为负数
复杂度：O(n!),空间O(n)
vector<vector<string> >res;
int _n;
vector<bool> is_col_chosen;
vector<bool> is_main_diag_chosen;
vector<bool> is_anti_diag_chosen;
bool is_chosen(int i, int j){
return is_col_chosen[j] || is_main_diag_chosen[i + j] || is_anti_diag_chosen[i - j + _n];
}
void dfs(int nth_queen, vector<int> &columns){
if(nth_queen == _n){
vector<string> one_res(_n, string(_n, '.'));
int row = 0;
for_each(one_res.begin(), one_res.end(), [&columns, &row](string &one_row){
one_row[columns[row++]] = 'Q';
});
res.push_back(one_res);
}
for(int j = 0; j < _n; ++j){
if(!is_chosen(nth_queen, j)){
is_col_chosen[j] = is_main_diag_chosen[nth_queen + j] = is_anti_diag_chosen[nth_queen - j + _n] = true;
columns.push_back(j);
dfs(nth_queen + 1,columns);
columns.pop_back();
is_col_chosen[j] = is_main_diag_chosen[nth_queen + j] = is_anti_diag_chosen[nth_queen - j + _n] = false;
}
}
}
vector<vector<string> > solveNQueens(int n){
_n = n;
is_col_chosen = is_main_diag_chosen = is_anti_diag_chosen = vector<bool>(2 * n, false);
vector<int> columns;
dfs(0, columns);
return res;
}
标签：style http color os io ar for div 问题
原文地址：http://blog.csdn.net/zhengsenlie/article/details/39177089
