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题目：
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.."] ]
代码：
class Solution { public: static vector<vector<string> > solveNQueens(int n) { vector<vector<string> > ret; if ( n==0 ) return ret; vector<string> tmp; vector<bool> colUsed(n,false); vector<bool> diagUsed1(2*n-1,false); vector<bool> diagUsed2(2*n-1,false); Solution::dfs(ret, tmp, n, colUsed, diagUsed1, diagUsed2); return ret; } static void dfs( vector<vector<string> >& ret, vector<string>& tmp, int n, vector<bool>& colUsed, vector<bool>& diagUsed1, vector<bool>& diagUsed2 ) { const int row = tmp.size(); if ( row==n ) { ret.push_back(tmp); return; } string curr(n,‘.‘); for ( size_t col = 0; col<n; ++col ) { if ( !colUsed[col] && !diagUsed1[col+n-1-row] && !diagUsed2[col+row] ) { colUsed[col] = !colUsed[col]; diagUsed1[col+n-1-row] = !diagUsed1[col+n-1-row]; diagUsed2[col+row] = !diagUsed2[col+row]; curr[col] = ‘Q‘; tmp.push_back(curr); Solution::dfs(ret, tmp, n, colUsed, diagUsed1, diagUsed2); tmp.pop_back(); curr[col] = ‘.‘; diagUsed2[col+row] = !diagUsed2[col+row]; diagUsed1[col+n-1-row] = !diagUsed1[col+n-1-row]; colUsed[col] = !colUsed[col]; } } } };
tips：
深搜写法：
1. 找到一个解的条件是tmp的长度等于n
2. 在一列中遍历每个位置，是否能够放置Q，并继续dfs；返回结果后，回溯tmp之前的状态，继续dfs。
一开始遗漏了对角线也不能在有超过一个Q的条件，补上之后就AC了。
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原文地址：http://www.cnblogs.com/xbf9xbf/p/4532922.html
