codechef Prime Distance On Tree FFT

先樹分治，對於每個點爲根處理一下，然後對這個點的子樹構成的多項式平方一下求一下素數個數即可。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<vector>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn=100005;
const double eps(1e-8);
typedef long long LL;
const double PI = acos(-1.0);
int isprim[maxn<<1];
void init(void){
    for(int i=2;i<2*maxn;i++){
        for(int j=2*i;j<2*maxn;j+=i){
            isprim[j]=1;
        }
    }
}
struct Complex
{
    double real, image;
    Complex(double _real, double _image)
    {
        real = _real;
        image = _image;
    }
    Complex(){}
};

Complex operator + (const Complex &c1, const Complex &c2)
{
    return Complex(c1.real + c2.real, c1.image + c2.image);
}

Complex operator - (const Complex &c1, const Complex &c2)
{
    return Complex(c1.real - c2.real, c1.image - c2.image);
}

Complex operator * (const Complex &c1, const Complex &c2)
{
    return Complex(c1.real*c2.real - c1.image*c2.image, c1.real*c2.image + c1.image*c2.real);
}

int rev(int id, int len)
{
    int ret = 0;
    for(int i = 0; (1 << i) < len; i++)
    {
        ret <<= 1;
        if(id & (1 << i)) ret |= 1;
    }
    return ret;
}

Complex A[maxn<<1];
void FFT(Complex* a, int len, int DFT)
{
    for(int i = 0; i < len; i++)
        A[rev(i, len)] = a[i];
    for(int s = 1; (1 << s) <= len; s++)
    {
        int m = (1 << s);
        Complex wm = Complex(cos(DFT*2*PI/m), sin(DFT*2*PI/m));
        for(int k = 0; k < len; k += m)
        {
            Complex w = Complex(1, 0);
            for(int j = 0; j < (m >> 1); j++)
            {
                Complex t = w*A[k + j + (m >> 1)];
                Complex u = A[k + j];
                A[k + j] = u + t;
                A[k + j + (m >> 1)] = u - t;
                w = w*wm;
            }
        }
    }
    if(DFT == -1) for(int i = 0; i < len; i++) A[i].real /= len, A[i].image /= len;
    for(int i = 0; i < len; i++) a[i] = A[i];
    return;
}
Complex a[maxn<<1];
int size[maxn];
int vis[maxn];
vector<int>g[maxn];
vector<int>gg,gg1;
int dp[maxn];
void dfs(int u,int pa){
    size[u]=1;
    dp[u]=0;
    gg.push_back(u);
    for(int v:g[u]){
        if(!vis[v]&&v!=pa){
            dfs(v,u);
            dp[u]=max(dp[u],size[v]);
            size[u]+=size[v];
        }
    }
}
void dfs2(int u,int pa,int d){
    gg.push_back(d);
    gg1.push_back(d);
    for(int v:g[u]){
        if(!vis[v]&&v!=pa){
            dfs2(v,u,d+1);
        }
    }
}
int num[maxn<<1];
LL xishu[maxn<<1];
LL get(vector<int>g2){
    int len=1,s=0;
    for(int q:g2){
        s=max(s,q);
    }
    while(len<=s) len=len<<1;
    len<<=1;
    for(int i=0;i<len;i++){
        num[i]=0;
    }
    for(int q:g2){
        num[q]++;
    }
    for(int i=0;i<len;i++){
        a[i]=Complex(num[i],0);
    }
    FFT(a,len,1);
    for(int i=0;i<len;i++){
        a[i]=a[i]*a[i];
    }
    FFT(a,len,-1);
    for(int i=0;i<len;i++){
        xishu[i]=(LL)(a[i].real+0.5);
    }
    for(int i=0;i<len/2;i++){
        xishu[i<<1]-=num[i];
    }
    LL ans=0;
    for(int i=2;i<len;i++){
        xishu[i]/=2;
        if(isprim[i]==0){
            ans+=xishu[i];
        }
    }
    return ans;
}
LL ans;
void solve(int u){
    gg.clear();
    for(int v:g[u]){
        if(!vis[v]){
            gg1.clear();
            dfs2(v,u,1);
            ans-=get(gg1);
        }
    }
    gg.push_back(0);
    ans+=get(gg);
}
void dfs1(int u){
    gg.clear();
    dfs(u,-1);
    int root=1,ans=100000000;
    for(int v:gg){
        dp[v]=max(dp[v],dp[u]-dp[v]);
        if(ans>dp[v]){
            ans=dp[v];
            root=v;
        }
    }
    vis[root]=1;
    solve(root);
    for(int v:g[root]){
        if(!vis[v]){
            dfs1(v);
        }
    }
}
int main()
{
    int n;
    init();
    cin>>n;
    for(int i=0;i<n-1;i++){
        int x,y;
        scanf("%d%d",&x,&y);
        g[x].push_back(y);
        g[y].push_back(x);
    }
    dfs1(1);
    double ans1=(1.0)*ans/((LL)n*(n-1)/2);
    printf("%.6lf\n",ans1);
}