BZOJ 1005 明明的烦恼

Description

自从明明学了树的结构,就对奇怪的树产生了兴趣...... 给出标号为1到N的点,以及某些点最终的度数,允许在任意两点间连线,可产生多少棵度数满足要求的树?

Input

第一行为N(0 < N < = 1000),接下来N行,第i+1行给出第i个节点的度数Di,如果对度数不要求,则输入-1

Output

一个整数,表示不同的满足要求的树的个数,无解输出0

Sample Input

3
1
-1
-1

Sample Output

2

HINT

两棵树分别为1-2-3;1-3-2

利用Purfer Sequence，参见：http://www.cnblogs.com/zhj5chengfeng/archive/2013/08/23/3278557.html

 #include<cstring>
 #include<cstdio>
 #include<cstdlib>
 using namespace std;

 #define maxn 1010
 int d[maxn],n,sum,cnt,tot,prime[maxn],num[maxn];
 bool exist[maxn];
 struct node
 {
     int a[maxn*],len;
     node(){memset(a,,sizeof(a)); len = ;}
     friend inline node operator *(node &x,int y)
     {
         node z; z.len = x.len + ;
         int i;
         for (i = ;i <= z.len;++i)
         {
             z.a[i] += x.a[i] * y;
             z.a[i+] += z.a[i] / ;
             z.a[i] %= ;
         }
         while (!z.a[z.len]) --z.len;
         return z;
     }

     inline void print() {for (int i = len;i;--i) printf("%d",a[i]);}
 }ans;

 inline void ready()
 {
     int i,j;
     for (i = ;i <= ;++i)
         if (!exist[i])
         {
             exist[i] = true;
             prime[++tot] = i;
             for (j = i*i;j <= ;j += i)
                 exist[j] = true;
         }
 }

 inline bool okay()
 {
     for (int i = ;i <= n;++i)
     {
         if (d[i] > ) sum += d[i] - ,++cnt;
         if (d[i] == ) return false;
     }
     if (sum + n - cnt > *(n-)) return false;
     if (cnt == n && sum != *(n-)) return false;
     return true;
 }

 inline void Div(int a,int bei)
 {
     if (a == ) return;
     if (bei == ) return;
     for (int i = ;i <= tot;++i)
     {
         if (a == ) break;
         if (a % prime[i] == )
         {
             int t = ;
             while (a % prime[i] == ) ++t,a /= prime[i];
             num[i] += t * bei;
         }
     }
 }

 inline void calc()
 {
     ready();
     for (int i = ;i <= n-;++i) Div(i,);
     Div(n - cnt,n-sum-);
     for (int i = ;i <= n--sum;++i) Div(i,-);
     for (int i = ;i <= n;++i) if (d[i] != -)
         for (int j = ;j < d[i];++j) Div(j,-);
     ans.a[] = ;
     for (int i = ;i <= tot;++i)
         for (int j = ;j <= num[i];++j)
             ans = ans * prime[i];
     ans.print();
 }

 int main()
 {
     scanf("%d",&n); int i;
     for (i = ;i <= n;++i) scanf("%d",d+i);
     if (!okay()) printf("%d",);
     else calc();
     return ;
 }