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

 1 #include<cstring>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 using namespace std;
 5 
 6 #define maxn 1010
 7 int d[maxn],n,sum,cnt,tot,prime[maxn],num[maxn];
 8 bool exist[maxn];
 9 struct node
10 {
11     int a[maxn*10],len;
12     node(){memset(a,0,sizeof(a)); len = 1;}
13     friend inline node operator *(node &x,int y)
14     {
15         node z; z.len = x.len + 3;
16         int i;
17         for (i = 1;i <= z.len;++i)
18         {
19             z.a[i] += x.a[i] * y;
20             z.a[i+1] += z.a[i] / 10;
21             z.a[i] %= 10;
22         }
23         while (!z.a[z.len]) --z.len;
24         return z;
25     }
26 
27     inline void print() {for (int i = len;i;--i) printf("%d",a[i]);}
28 }ans;
29 
30 inline void ready()
31 {
32     int i,j;
33     for (i = 2;i <= 1000;++i)
34         if (!exist[i])
35         {
36             exist[i] = true;
37             prime[++tot] = i;
38             for (j = i*i;j <= 1000;j += i)
39                 exist[j] = true;
40         }
41 }
42 
43 inline bool okay()
44 {
45     for (int i = 1;i <= n;++i)
46     {
47         if (d[i] > 0) sum += d[i] - 1,++cnt;
48         if (d[i] == 0) return false;
49     }
50     if (sum + n - cnt > 2*(n-1)) return false;
51     if (cnt == n && sum != 2*(n-1)) return false;
52     return true;
53 }
54 
55 inline void Div(int a,int bei)
56 {
57     if (a == 0) return;
58     if (bei == 0) return;
59     for (int i = 1;i <= tot;++i)
60     {
61         if (a == 1) break;
62         if (a % prime[i] == 0)
63         {
64             int t = 0;
65             while (a % prime[i] == 0) ++t,a /= prime[i];
66             num[i] += t * bei;
67         }
68     }
69 }
70 
71 inline void calc()
72 {
73     ready();
74     for (int i = 1;i <= n-2;++i) Div(i,1);
75     Div(n - cnt,n-sum-2);
76     for (int i = 1;i <= n-2-sum;++i) Div(i,-1);
77     for (int i = 1;i <= n;++i) if (d[i] != -1)
78         for (int j = 1;j < d[i];++j) Div(j,-1);
79     ans.a[1] = 1;
80     for (int i = 1;i <= tot;++i)
81         for (int j = 1;j <= num[i];++j)
82             ans = ans * prime[i];
83     ans.print();
84 }
85 
86 int main()
87 {
88     scanf("%d",&n); int i;
89     for (i = 1;i <= n;++i) scanf("%d",d+i);
90     if (!okay()) printf("%d",0);
91     else calc();
92     return 0;
93 }
View Code

 

posted @ 2015-01-09 13:15  lmxyy  阅读(209)  评论(0)  编辑  收藏  举报