bzoj 3451: Tyvj1953 Normal [fft 点分治 期望]

3451: Tyvj1953 Normal

题意：

N 个点的树，点分治时等概率地随机选点，代价为当前连通块的顶点数量，求代价的期望值

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百年难遇的点分治一遍AC!!!

今天又去翻了一下《具体数学》上的离散概率，对期望有了一点新认识吧。

本题根据期望的线性性质，计算每个点的代价期望加起来。

一个点v产生了代价，它在u选为中心时所在的cc里，并且(u,v)路径上没有其他点已经被选择。概率为\(\frac{1}{(u,v)之间包含u,v点的个数}\)

统计每种长度的路径有多少个

点分治+生成函数统计就好了

值得注意的是，要用总-同一个子树的做法，否则fft的次数界会与最大子树深度相关，可以被卡成n^2

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <ctime>
using namespace std;
typedef long long ll;
const int N = (1<<16) + 5, M = 1e5+5;
const double PI = acos(-1.0);
inline int read() {
    char c=getchar(); int x=0,f=1;
    while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
    return x*f;
}

struct meow{
    double x, y;
    meow(double a=0, double b=0):x(a), y(b){}
};
meow operator +(meow a, meow b) {return meow(a.x+b.x, a.y+b.y);}
meow operator -(meow a, meow b) {return meow(a.x-b.x, a.y-b.y);}
meow operator *(meow a, meow b) {return meow(a.x*b.x-a.y*b.y, a.x*b.y+a.y*b.x);}
meow conj(meow a) {return meow(a.x, -a.y);}
typedef meow cd;

namespace fft {
	int maxlen = 1<<16, rev[N];
	cd omega[N], omegaInv[N];
	void init(int lim) {
		maxlen = 1; while(maxlen < lim) maxlen<<=1;
		for(int i=0; i<maxlen; i++) {
			omega[i] = cd(cos(2*PI/maxlen*i), sin(2*PI/maxlen*i));
			omegaInv[i] = conj(omega[i]);
		}
	}
	void dft(cd *a, int n, int flag) {
		cd *w = flag == 1 ? omega : omegaInv;
		int k = 0; while((1<<k) < n) k++;
		for(int i=0; i<n; i++) {
			rev[i] = (rev[i>>1]>>1) | ((i&1)<<(k-1));
			if(i < rev[i]) swap(a[i], a[rev[i]]);
		}
		for(int l=2; l<=n; l<<=1) {
			int m = l>>1;
			for(cd *p = a; p != a+n; p += l) 
				for(int k=0; k<m; k++) {
					cd t = w[maxlen/l*k] * p[k+m];
					p[k+m] = p[k] - t;
					p[k] = p[k] + t;
				}
		}
		if(flag == -1) for(int i=0; i<n; i++) a[i].x /= n;
	}
	void sqr(cd *a, int n) {
		dft(a, n, 1);
		for(int i=0; i<n; i++) a[i] = a[i] * a[i];
		dft(a, n, -1);
	}
}

int n;
struct edge{int v, ne;} e[N<<1];
int cnt, h[N];
inline void ins(int u, int v) {
	e[++cnt] = (edge){v, h[u]}; h[u] = cnt;
	e[++cnt] = (edge){u, h[v]}; h[v] = cnt;
}
int rt, f[N], vis[N], size[N], all;
void dfs_rt(int u, int fa) {
	size[u] = 1; f[u] = 0;
	for(int i=h[u];i;i=e[i].ne) {
		int v = e[i].v;
		if(vis[v] || v == fa) continue;
		dfs_rt(v, u);
		size[u] += size[v];
		f[u] = max(f[u], size[v]);
	}
	f[u] = max(f[u], all-size[u]);
	if(f[u] < f[rt]) rt = u;
}

int d[N], c[N], lim;
void dfs_get(int u, int fa) {
	for(int i=h[u];i;i=e[i].ne) {
		int v = e[i].v;
		if(vis[v] || v == fa) continue;
		d[v] = d[u] + 1; c[d[v]]++;
		lim = max(lim, d[v]+1);
		dfs_get(v, u);
	}
}

cd a[N];
int g[N], t[N];
void cal(int lim, int flag) { 
	int n = 1; while(n<lim<<1) n<<=1; //printf("cal------- %d  %d   %d\n", lim, n, flag);
	//for(int i=0; i<lim; i++) printf("%d ", c[i]); puts(" c");
	for(int i=0; i<lim; i++) a[i] = cd(c[i]), c[i] = 0;
	for(int i=lim; i<n; i++) a[i] = cd();
	fft::sqr(a, n);
	//for(int i=0; i<lim<<1; i++) printf("%.0lf ", a[i].x); puts(" a");
	for(int i=1; i<lim<<1; i++) t[i] += flag * (int) floor(a[i-1].x + 0.5);
	//for(int i=0; i<lim<<1; i++) printf("%d ", t[i]); puts(" t");
}

void dfs(int u) { //printf("\n-----------dfs---------- %d\n", u);
	vis[u] = 1;
	d[u]=0; c[0]++; lim=1; dfs_get(u, 0); 
	for(int i=0; i<lim<<1; i++) t[i] = 0;
	cal(lim, 1);
	int _lim = lim;
	for(int i=h[u];i;i=e[i].ne) {
		int v = e[i].v;
		if(vis[v]) continue;
		d[v] = 1; c[1]++; lim = 2; dfs_get(v, u);
		cal(lim, -1);
	}
	for(int i=0; i<_lim<<1; i++) g[i] += t[i];// printf("final %d  %d\n", i, t[i]);

	for(int i=h[u];i;i=e[i].ne) {
		int v = e[i].v;
		if(vis[v]) continue;
		all = size[v]; rt=0; dfs_rt(v, u);
		dfs(rt);
	}
}

int main() {
	freopen("in", "r", stdin);
	n=read();
	fft::init(n<<1);
	for(int i=1; i<n; i++) ins(read()+1, read()+1);
	all=n; rt=0; f[0]=1e9; dfs_rt(1, 0);
	dfs(rt);
	double ans = 0;
	//for(int i=1; i<=n; i++) printf("%d ", g[i]); puts("");
	for(int i=1; i<=n; i++) if(g[i]) ans += (double) g[i] / i;
	printf("%.4lf", ans);
}