HDU2553 n Queen question "backtracking method"

N Queen's question

Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total Submission (s): 9579 Accepted Submission (s): 4314
Problem Description
N-Queens are placed in the N*n's checkered checkerboard, making them not attack each other (that is, any 2 queens are not allowed to be in the same row, in the same column, or in a diagonal line with a 45-angle checkerboard border.)
Your task is to find out how many legal placement methods are available for a given n.

Input
There are several lines, one positive integer n≤10 per line, indicating the number of boards and queens, or, if n=0, the end.

Output
A number of rows, one positive integer per line, representing the number of different placements of the queen corresponding to the input row.

Sample Input
1
8
5
0

Sample Output
1
92

10

Title: N-Queens on the N*n board, n Queens cannot appear on the same line, in the same column, or in the same diagonal line.

Idea: Direct enumeration judgment is too slow, considering that each row has only one queen per column, then using an array of c[x] represents the first

The column number where the Queen is placed on the X row, that is, X for the row, and C[x] for the column. Judging whether the clash with the Queen of the Front can

Determines whether the current line cur is in conflict with the previous 0~j line.

C[cur] = = C[j] | | Cur-c[cur] = = J-c[j] | | Cur+c[cur] = = J+c[j] respectively determine whether in the same column, the same

The main diagonal, on the same diagonal.

However, you can continue to optimize. Directly with a vis[3][?] Directly determine if the column and diagonal of the currently trying queen are

There are other queens, that is,!vis[0][i] in the column whether there are other queens,!vis[1][cur+i] where the vice-diagonal has other royal

!vis[2][cur-i+n] There are other queens on the sub-diagonal.

But this commits the code also timed out ...

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring>using namespace Std;int vis[3][50],c[30],tot,n;void Search (int cur) {    if (cur = = N)    {        tot++;    }    else    {for        (int i = 0; i < N; i++)        {            if (!vis[0][i] &&!vis[1][cur+i] &&!vis[2][cur-i+n ])            {                Vis[0][i] = vis[1][cur+i] = Vis[2][cur-i+n] = 1;                Search (cur+1);                Vis[0][i] = Vis[1][cur+i] = Vis[2][cur-i+n] = 0;}}}    int main () {    while (~scanf ("%d", &n) && N)    {        tot = 0;        memset (vis,0,sizeof (Vis));        Search (0);        printf ("%d\n", tot);    }    return 0;}

Make a decisive statement

#include <iostream>using namespace Std;int main () {    int a[23] = {0,1,0,0,2,10,4,40,92,352,724};    int N;    while (CIN >> N)    {        if (n==0) break            ;        else            cout << a[n] << Endl;    }}

HDU2553 n Queen question "backtracking method"