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..”]
]
題意：對於一個nxn的棋盤，裏面放n個皇后，彼此之間不再同一行同一列以及同一條對角線上，皇后用Q代替，空格用.代替，求出所有的可能結果。
解題思路：vector < vector < int > > rt表示棋盤，vector< int > v表示n個皇后，v[i]表示i行的皇后所在的列，對i從0到n依次進行嘗試，看是否存在解。
bool iv(vector<int> v,int k)
{
for(int i=1;i<k;++i)
{
if(v[i]==v[k])
{
return false;
}
double xt=k-i;
double yt=v[i]-v[k];
double xy=yt/xt;
if(xy==1||xy==-1)
{
return false;
}
}
return true;
}
void sq(int k,int n,vector<int>& v,vector<vector<string>>& rt)
{
for(int j=1;j<=n;++j)
{
v[k]=j;
if(iv(v,k))
{
if(k==n)
{
vector<string> rv;
for(int i=1;i<=n;++i)
{
string t;
for(int ii=1;ii<v[i];++ii)
{
t+=".";
}
t+="Q";
for(int ii=v[i]+1;ii<=n;++ii)
{
t+=".";
}
rv.push_back(t);
}
rt.push_back(rv);
}
else
{
sq(k+1,n,v,rt);
}
}
else
{
v[k]=0;
}
}
}
vector<vector<string>> solveNQueens(int n) {
if(n<1)
{
return vector<vector<string>>();
}
vector<int> v(n+1,0);
vector<vector<string>> ret;
sq(1,n,v,ret);
return ret;
}