【bzoj1211】 HNOI2004—树的计数

http://www.lydsy.com/JudgeOnline/problem.php?id=1211 (题目链接)

题意

　　一个有n个结点的树，设它的结点分别为v1, v2, …, vn，已知第i个结点vi的度数为di，问满足这样的条件的不同的树有多少棵。给定n，d1, d2, …, dn，编程需要输出满足d(vi)=di的树的个数。

Solution

　　prufer序列，明明的烦恼简化版。

代码

// bzoj1211
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<map>
#define inf 2147483640
#define LL long long
#define free(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout);
using namespace std;
inline LL getint() {
    LL x=0,f=1;char ch=getchar();
    while (ch>'9' || ch<'0') {if (ch=='-') f=-1;ch=getchar();}
    while (ch>='0' && ch<='9') {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
 
const int maxn=200;
int d[maxn],a[maxn],np[maxn],n;
 
void pls(int x,int f) {
    for (int i=2;i<=x;i++)
        if (!np[i]) for (int j=i;j<=x;j*=i) a[i]+=f*x/j;
}
int main() {
    np[1]=1;
    for (int i=2;i<=150;i++)
        if (!np[i]) for (int j=i*2;j<=150;j+=i) np[j]=1;
    scanf("%d",&n);
    int sum=0;
    if (n==1) {
        int x;scanf("%d",&x);
        if (!x) printf("1");
        else printf("0");
        return 0;
    }
    for (int i=1;i<=n;i++) {
        scanf("%d",&d[i]);
        if (!d[i]) {printf("0");return 0;}
        d[i]--;
        sum+=d[i];
        pls(d[i],-1);
    }
    if (sum!=n-2) {printf("0");return 0;}
    LL ans=1;
    pls(sum,1);
    for (int i=2;i<=150;i++)
        for (int j=1;j<=a[i];j++) ans*=(LL)i;
    printf("%lld",ans);
    return 0;
}