HDU 2553 n皇后问题(回溯法)

 1 #include<iostream>
 2 #include<cmath>
 3 using namespace std;
 4 const int maxn=11;
 5 int b[maxn],a[maxn],sum,n;
 6 
 7 void dfs(int cur)
 8 {
 9     if(cur == n+1)//递归边界，就有一种摆法
10         sum++;
11     else
12         for(int j = 1; j <=n; j++)
13         {
14             int ok=1;
15             a[cur] = j;//尝试把第cur行的皇后放在第j列
16             for(int i = 1; i<cur; i++)   //检查是否和前面的皇后冲突
17                 if(a[i] == a[cur] || abs(i - cur) == abs(a[i] - a[cur]))
18                 {
19                     ok=0;
20                     break;
21                 }
22             if(ok)
23                 dfs(cur+1);//如果合法，继续递归
24         }
25 }
26 
27 int main()
28 {
29     for(int i = 1; i <=maxn; i++)
30     {
31         sum = 0;
32         n= i;
33         dfs(1);
34         b[i] = sum;
35     }
36      while(cin>>n && n)
37         cout<<b[n]<<endl;
38     return 0;
39 }

1 #include <cstdio>
2 main()
3 {
4     int n,a[13]={0,1,0,0,2,10,4,40,92,352,724,2680,14200};
5     while(scanf("%d",&n))
6         printf("%d\n",a[n]);
7 }