Va 750-8 queens chess problem

Question: Calculate the eight queens that must pass through the designated time.

Analysis: search. Because eight queens only have 92 groups of solutions, they can calculate 92 groups of solutions directly, and then query and output the solutions.

Here I use bitwise operations to calculate the eight queens and reduce the amount of code.

First, consider the impact of a Queen. The relationship between the next attack and the last attack is as follows:

A queen will influence two oblique directions (from top to bottom );

The left oblique influences the next layer to move one to the left, and the right to move one to the right;

Therefore, we use bitwise representation for the three effects,

L, m, and R are in three situations, respectively. The bit representation of all the points that the Queen was able to attack;

If the I-th element is used, the three elements correspond to the following places:

L = (L | (1 <I) <1, M | (1 <I), (R | (1 <I)> 1;

Here I am the selected enumeration element. I can use the bit operation to find the final Available Bit (-P & P );

But it is troublesome to expand (because it is 2 ^ I, not I ).

Note: There are % 2D output numbers.

#include <iostream>#include <cstdlib>#include <cstdio>using namespace std;int  ans[92][9],temp[9];int  Count = 0;void dfs(int L, int M, int R, int d){if (d == 8) {Count ++;return;}for (int i = 0 ; i < 8 ; ++ i) {if (((L|M|R)&(1<<i)) == 0) {for (int j = 0 ; j < d ; ++ j)temp[j] = ans[Count][j];ans[Count][d] = i+1;dfs((L|(1<<i))<<1, M|(1<<i), (R|(1<<i))>>1, d+1);for (int j = 0 ; j < d ; ++ j)ans[Count][j] = temp[j];}}}int main(){Count = 0;dfs(0, 0, 0, 0);int t,x,y;while (~scanf("%d",&t)) while (t --) {scanf("%d%d",&x,&y);printf("SOLN       COLUMN\n");printf(" #      1 2 3 4 5 6 7 8\n\n");int count = 1;for (int i = 0 ; i < Count ; ++ i)if (ans[i][y-1] == x) {printf("%2d     ",count ++);for (int j = 0 ; j < 8 ; ++ j)printf(" %d",ans[i][j]);printf("\n");}if (t) printf("\n");}return 0;}

Va 750-8 queens chess problem