N皇后問題

轉自：http://www.cnblogs.com/chuanlong/archive/2013/04/21/3033471.html

Problem Description

在N*N的方格棋盤放置了N個皇后，使得它們不相互攻擊（即任意2個皇后不允許處在同一排，同一列，也不允許處在與棋盤邊框成45角的斜線上。
你的任務是，對於給定的N，求出有多少種合法的放置方法。

 

Input

共有若干行，每行一個正整數N≤10，表示棋盤和皇后的數量；如果N=0，表示結束。

 

Output

            共有若干行，每行一個正整數，表示對應輸入行的皇后的不同放置數量。

 

Sample Input

1
8
5
0

 

Sample Output

1
92
10

 

方法一：遞歸  容易理解，但效率低點，也能Accepted

<span style="font-size:14px;">#include <stdio.h>  
#include <math.h>
#include <stdlib.h>
int q[11];  //q[1] means the coordinate of queue is (1, q[1])
int result[11];//to save the time, so record the previous time
int n;
int check(int k) //check if the kth queen is conflict with previous ones
{
    int i;
    i = 1;
    while (i < k)
    {
        //k - i == abs(q[k] - q[i]) means the current queen is not 45° with other queens
        if (q[i] == q[k] || k - i == abs(q[k] - q[i]))
            return 0;
        i++;
    }
    return 1;
}
int count;  //record the count of rank
int flag;
void DFS(int step)
{
    int i, j, k;
    if (step == n + 1)
        count++; <span style="color:#3333ff;">   //需要打印具體的放置方式時，可以在這裏打印，q[i]表示第i個皇后放在第i行第q[i]列</span>
    else
    {
        for (i = 1; i <= n; i++)
        {
            q[step] = i;
            if (check(step) == 0)
                continue;
            DFS(step + 1);
        }
    }
}
int main()  
{  
    while (scanf("%d", &n) != EOF && n)
    {
        //memset();
        count = 0;
        if (result[n] == 0)
        {
            DFS(1);
            result[n] = count;
        }
        printf("%d\n", result[n]);
    }
    return 0;  
}</span>

方法二：運行效率更快，但不好理解

<span style="font-size:14px;">#include <stdio.h>  
#include <math.h>
int q[11];  //q[1] means the coordinate of queue is (1, q[1])
int result[11];//to save the time, so record the previous time
int check(int k) //check if the kth queen is conflict with previous ones
{
    int i;
    i = 1;
    while (i < k)
    {
        //k - i == abs(q[k] - q[i]) means the current queen is not 45° with other queens
        if (q[i] == q[k] || k - i == abs(q[k] - q[i]))
            return 0;
        i++;
    }
    return 1;
}
//begin with the first queen, find the position for first queen
//and use DFS to go on to find next queen's position, if u can't find it, u can go back to change the previous queen's position
//untill there is no position for the fist queue to change 
int Nqueens(int n)
{
    int count, i, j, k;
    count = 0;  //record the count of rank
    //begin with first queen and the zero position
    k = 1;
    q[k] = 0;
    while (k > 0)
    {
        q[k]++;
        //when q[k] <= n, u can go on to find q[k] satisfied the condition
        while (q[k] <= n && check(k) == 0)
            q[k]++;
        //if q[k] <= n, which means u can find position for current queen
        if (q[k] <= n)
        {
            //means u find the last queen, so u can record the counr
            if (k == n)
                count++;
            //if it's not the last queen, u should go on to find the place of next queen
            //and for next queen, u should begin with zero.
            else
            {
                k++;
                q[k] = 0;
            }
        }
        else
            //if u can't find the position for current queen, u should go back to modify the previous queen's position
            k--;
    }
    return count;
}
int main()  
{  
    int n;
    while (scanf("%d", &n) != EOF && n)
    {
        if (result[n] == 0)
            result[n] = Nqueens(n);
        printf("%d\n", result[n]);
    }
    return 0;  
}</span>