[Leetcode series] recursive solution to n queen problem-bit Operation Method

N queen's question:

Given the 8*8 board, place n Queens so that they cannot attack each other (that is, the two queens cannot be placed on the same row/column/positive/negative diagonal). How many placement methods are there to solve?

There are a lot of answers to this question on the Internet, but I don't see much of the bit operation solution. I will not repeat the code and illustration below.

1 class solution {2 public: 3/* uses the backtracking algorithm implemented by bit operations. scan by row to detect columns that can be placed. 4 * 'limit'-all are '1 '. it indicates that all columns are occupied by 5 * 'h', which is the vertical projection of all current Queen columns on the row. if H = limit, the search is complete. Answer ++. 6 * 'r'-it is the vertical projection of all the Queen's diagonal lines. 7 * 'l'-it is the vertical projection of the diagonal lines of all queens currently. 8 * H | r | L-all occupied places. therefore, Pos = Limit &(~ (H | r | L) is all free positions. 9 * P = POS & (-Pos) Find the rightmost 1. pos-= P indicates placing a queen on the one represented by P. 10 * 'H + p'-is the vertical projection of the new queen's columns on the row. 11 * '(R + p) <1' is the vertical projection of the new queen's diagonal. because we move to the next row, the projection is tilted from right to left, 12 * we need to move to the next row and then shift to the left position. 13 * '(L + p)> 1' is the vertical projection of the new queen's diagonal line. because we move to the next row, the projection is tilted from left to right, and 14 * We need to translate the position to the right after moving to the next row. 15 */16 int ans, limit; 17 int totalnqueens (int n) {18 ans = 0; 19 limit = (1 <n)-1; 20 DFS (0, 0, 0); 21 return ans; 22} 23 void DFS (in T h, int R, int L) {24 if (H = Limit) {25 ans ++; 26 return; 27} 28 int Pos = Limit &(~ (H | r | L); // use a bit and remove a high value of 0 to obtain all positions that can be placed in this row. 29 While (POS) {30 // because the parts with more than 8 POS bits are 0, the following analysis only targets low 8-bit 31 int P = POS & (-Pos ); // The result of obtaining a negative number is the POs rounded up to + 1, and the & Pos result is the lowest position of the POs. 1 32 pos-= P; // remove this 33 DFS (H + P, (R + p) <1, (L + p)> 1 ); // Add the new p34 directly to the column direction restriction. // For The Next row, the diagonal limit is one grid 35} 36} 37 on both sides of P };