IT company interview questions collected-C-related-eight queens explanation

Some people may not be able to read the previous post. I made a special arrangement and tried to make it easy to understand. The comments are very detailed and are basically annotated.

First, paste a picture. It is easier to see the picture.

Paste detailed code

# Include <stdio. h> # define row 8 // represents the column, the coordinate is X # define Col 8 // represents the row, and the coordinate is y # define num 7 // problem, array subscript 0-7 # define true 0 // true # define false 1 // false # define Yes 1 // queen # define no 0 // no queen # define no_back 0/ /indicates that backtracking is not required # define back 1 // indicates that backtracking is required for int A [row] [col]; // The default value is 0, indicating that the coordinate has no queen. If it is changed to 1, it indicates that the coordinate has a queen int K = 0; // indicates the condition in K. // n indicates the subscript of the column to be traversed, that is, the value of X. Rowint queens (int n) {int I, j; // output eight queens board if (n> num) {printf ("---------- % 06d ---------- \ n ", ++ K); for (I = 0; I <row; I ++) {for (j = 0; j <Col; j ++) printf ("% s", a [I] [J] = 1? "○": "●"); // 0 indicates that it is not a queen. 1 indicates that it is a queen printf ("\ n ");} printf ("-------------------------- \ n"); return back; // indicates normal return and backtracking} // If the traversal is not complete, for (j = 0; j <Col; j ++) // traverse the Y value 0-> Col {int flag = true; // This loop traverses the X value, that is, the Y value remains unchanged, X value variation for (I = 0; I <n; I ++) // Based on the passed column subscript x value, row value, traverse {// if there is a queen in the left corner, Northeast southwest direction, right corner, and northwest southeast direction, this cycle will be launched, perform the next loop if (a [I] [J] = Yes | (I + J-N)> = 0 & A [I] [I + J-N] = Yes) | (J + n-I <col) & A [I] [J + n-I] = Yes) {flag = false; break ;}} if (flag = true) // if the previous cycle does not show this situation {// place queen a [n] [J] = Yes here; if (Queens (n + 1) = back) {// if the lower-layer traversal requires backtracking a [n] [J] = no; // do not place the queen here} else {// if the lower-layer traversal does not require backtracking, return no_back ;}}// when traversing the Y value, there is no suitable one, and flag = false; it indicates that the lower-level cannot place the queen, return back to return back;} int main (void) {queens (0); Return 0 ;}

There is a single-bit operation solution, which is also very good. If you are interested, please take a look. Http://blog.csdn.net/wang6279026/article/details/8661904