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.."] ]
class Solution {
vector<bool> row,col,lslash,rslash;
vector<int> pos;
vector<vector<string>> result;
public:
void dfs(int deep,int n){
if(deep==n){
vector<string> ans;
for(int i = 0; i < n; i++)
{
string s;
for(int j = 0; j < pos[i]; j++)
s += ‘.‘;
s += ‘Q‘;
for(int j = 0; j < n - (pos[i] + 1); j++)
s += ‘.‘;
ans.push_back(s);
}
result.push_back(ans);
return;
}
for(int i=0;i<n;i++){
if(!row[deep]&&!col[i]&&!lslash[deep+i]&&!rslash[deep+n-1-i]){
row[deep]=col[i]=lslash[deep+i]=rslash[deep+n-1-i]=true;
pos[deep]=i;
dfs(deep+1,n);
row[deep]=col[i]=lslash[deep+i]=rslash[deep+n-1-i]=false;
}
}
}
vector<vector<string> > solveNQueens(int n) {
row.assign(n,false);//某行是否有皇后了
col.assign(n,false);//某列是否有皇后了
lslash.assign(2*n,false);//左斜线是否有皇后了
rslash.assign(2*n,false);//右斜线是否有皇后了
pos.assign(n,0);
dfs(0,n);
return result;
}
};原文：http://blog.csdn.net/starcuan/article/details/18862175