暑假集训Day14 F (树上期望)

题目链接在这里：Problem - F - Codeforces

由于期望的可加性，我们可以单独考虑每一个点对整体期望的贡献值，对于一个点，如果它被删掉了，一定是从根节点到它本身这条路径上有点被删了，我们考虑的是这个点对整体的贡献，所以假设它的深度为d，贡献值即为1*(1/d)+0*((d-1)/d)，花见一下就是1/d

#include "bits/stdc++.h"
using namespace std;
const int MAX=1e5+5;
int n,cnt,d[MAX],num[MAX],nu;
vector <int> adj[MAX];
double f[MAX],ans;
int main(){
    freopen ("f.in","r",stdin);
    freopen ("f.out","w",stdout);
    int i,j,u,v,zt,now;
    vector <int> x;
    scanf("%d",&n);
    for (i=1;i<n;i++){
        scanf("%d%d",&u,&v);
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    queue <int> q;
    while (!q.empty()) q.pop();
    q.push(1);now=1;
    memset(f,0,sizeof(f));
    f[1]=1;
    while (!q.empty()){
        zt=q.front();q.pop();
        for (i=1;i<=adj[zt].size();i++)
            f[now+i]=1.0/(double)(now+i)+(double)(f[now+i-1]+1)*(double)(i)/(double)(now+i);
        now+=adj[zt].size();
    }
    memset(d,0,sizeof(d));
    d[1]=1;q.push(1);
    memset(num,0,sizeof(num));
    nu=0;
    while (!q.empty()){
        zt=q.front();q.pop();
        nu=max(nu,d[zt]);
        num[d[zt]]++;
        //cout<<zt<<' '<<d[zt]<<' '<<num[d[zt]]<<endl;
        for (i=0;i<adj[zt].size();i++){
            if (d[adj[zt][i]]!=0) continue;
            d[adj[zt][i]]=d[zt]+1;
            q.push(adj[zt][i]);
            //cout<<"fuck "<<zt<<' '<<adj[zt][i]<<endl;
        }
    }
    //cout<<nu<<endl;
    for (i=1;i<=nu;i++){
        //cout<<num[i]<<' ';
        ans+=(double)num[i]/(double)i;
    }//cout<<endl;
    printf("%.8lf",ans);
    return 0;
}