【N-Quens II】cpp

题目：

Follow up for N-Queens problem.

Now, instead outputting board configurations, return the total number of distinct solutions.

代码：

class Solution {
public:
        int totalNQueens(int n)
        {
            int ret = 0;
            if ( n==0 ) return ret;
            vector<bool> colUsed(n,false), diagUsed1(2*n-1,false), diagUsed2(2*n-1,false);
            Solution::dfs(ret, 0, n, colUsed, diagUsed1, diagUsed2);
            return ret;
        }
        static void dfs( int &ret, int row, int n, vector<bool>& colUsed, vector<bool>& diagUsed1, vector<bool>& diagUsed2 )
        {
            if ( row==n ) { ret++; return; }
            for ( size_t col = 0; col<n; ++col ){
                if ( !colUsed[col] && !diagUsed1[col+n-1-row] && !diagUsed2[col+row] )
                {
                    colUsed[col] = diagUsed1[col+n-1-row] = diagUsed2[col+row] = true;
                    Solution::dfs(ret, row+1, n, colUsed, diagUsed1, diagUsed2);
                    diagUsed2[col+row] = diagUsed1[col+n-1-row] = colUsed[col] = false;
                }
            }
        }
};

tips：

如果还是用深搜的思路，这个N-Queens II要比N-Queens简单一些，可能是存在其他的高效解法。

===========================================

第二次过这道题，经过了N-Queens，这道题顺着dfs的思路就写下来了。

class Solution {
public:
        int totalNQueens(int n)
        {
            int ret = 0;
            if ( n<1 ) return ret;
            vector<bool> colUsed(n, false);
            vector<bool> r2l(2*n-1, false);
            vector<bool> l2r(2*n-1, false);
            Solution::dfs(ret, n, 0, colUsed, r2l, l2r);
            return ret;
        }
        static void dfs( 
            int& ret,  
            int n,
            int index,
            vector<bool>& colUsed,
            vector<bool>& r2l,
            vector<bool>& l2r)
        {
            if ( index==n )
            {
                ret++;
                return;
            }
            for ( int i=0; i<n; ++i )
            {
                if ( colUsed[i] || r2l[i-index+n-1] || l2r[i+index] ) continue;
                colUsed[i] = true;
                r2l[i-index+n-1] = true;
                l2r[i+index] = true;
                Solution::dfs(ret, n, index+1, colUsed, r2l, l2r);
                colUsed[i] = false;
                r2l[i-index+n-1] = false;
                l2r[i+index] = false;
            }
        }
};