（赛前练手 #10）BZOJ1005 [HNOI2008]明明的烦恼（prufer数列 + 高精）

1005: [HNOI2008]明明的烦恼
Time Limit: 1 Sec Memory Limit: 162 MB
Submit: 6664 Solved: 2626
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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

Source

先粘一波其它博主的blog吧…
https://blog.csdn.net/jiangshibiao/article/details/22651081
其实我们还是利用prufer数列与树的一一对应关系，然后我们就可以利用排列组合来求了
AC Code:

#include
#define rg register
#define il inline
#define maxn 3000010
#define ll long long
#define eps 1e-8
using namespace std;
il int read(){rg int x = 0 , w = 1 ; rg char ch = getchar();while (ch < '0' || ch > '9'){if (ch == '-') w = -1; ch = getchar();}while (ch >= '0' && ch <= '9'){x = (x << 3) + (x << 1) + ch - '0' ; ch = getchar();}return x * w;}
int q[maxn << 1] , tot , rnx[maxn] , a[maxn] , b[maxn] , c[maxn] , val[maxn] , primes[maxn];
int flag[maxn] , p[maxn] , delta;
bool vis[maxn];
void prime(int n){
	vis[1] = 1;
	for (rg int i = 2 ; i <= n ; ++i){
		if (!vis[i]) {
			primes[++tot] = i;
			rnx[i] = tot;
		}
		for (rg int j = 1 ; j <= tot && i * primes[j] <= n; ++j){
			vis[i * primes[j]] = 1;
			if (!(i % primes[j])) break;	
		}
	}	
}
void fenjie(int x){
	delta += p[x];
	rg int tmp = x , i = 1;
	if (!vis[x]){
		val[rnx[x]] += delta;
		return;
	}
	while (vis[tmp] && tmp != 1){
		while (!(tmp % primes[i])){
			val[i] += delta;
			tmp /= primes[i];
		}
		++i;
	}
	if (tmp != 1)
		val[rnx[tmp]] += delta;
}
int main(){
	prime(500000);
	int n = read() , x , res = 0 , cnt = 0 , sum = 0;
	for (rg int i = 1 ; i <= n ; ++i){
		x = read();
		if (x > 0){
			q[++cnt] = x;
			sum += x - 1;
		}
		else if (x == 0){
			puts("0");
			return 0;
		}
		else ++res;
	}
	if (sum > (2 * n - 2)){
		puts("0");
		return 0;	
	}
	rg int maxr = -1;
	maxr = max(maxr , n - 1);
	p[1]++;
	p[n - 1]--;
	p[1]--;
	p[n - 1 - sum]++;
	for (rg int i = 1 ; i <= cnt ; ++i){
		p[1]--;
		p[q[i]]++;
		maxr = max(q[i] , maxr);	
	}
	p[res] += n - 2 - sum;
	p[res + 1] -= n - 2 - sum;
	maxr = max(res + 1 , maxr);
	for (rg int i = 1 ; i <= 200005; ++i){
		fenjie(i);	
	}
	rg int len = 1;
	a[len] = 1;
	for (rg int i = 1 ; i <= 1000 ; ++i){
		if (!val[i]) continue;
		for (rg int j = 1 ; j <= val[i] ; ++j){
			int xx = primes[i];
			x = 0;
			for (rg int k = 1 ; k <= len ; ++k){
				a[k] = a[k] * xx + x;
				x = a[k] / 10;
				a[k] = a[k] % 10;
			}
			while(x > 0){
				a[++len] = x % 10;
				x /= 10;	
			}
		}
	}
	for (rg int i = len ; i >= 1 ; --i)
		printf("%d" , a[i]);
	return 0;
}