【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。
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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; } } };