bzoj4675(点分治)

题目链接

https://www.lydsy.com/JudgeOnline/problem.php?id=4675

题解

首先显然这三个人的期望是相互独立的,然后对每个能得分的点对来说(假设他能选 $k$ 个点),他产生的贡献是

那么只要统计一下能得分的点数就可以了,这个直接点分就行。。




代码

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/**
*         ┏┓    ┏┓
*         ┏┛┗━━━━━━━┛┗━━━┓
*         ┃       ┃  
*         ┃   ━    ┃
*         ┃ >   < ┃
*         ┃       ┃
*         ┃... ⌒ ...  ┃
*         ┃ ┃
*         ┗━┓ ┏━┛
*          ┃ ┃ Code is far away from bug with the animal protecting          
*          ┃ ┃ 神兽保佑,代码无bug
*          ┃ ┃           
*          ┃ ┃       
*          ┃ ┃
*          ┃ ┃           
*          ┃ ┗━━━┓
*          ┃ ┣┓
*          ┃ ┏┛
*          ┗┓┓┏━━━━━━━━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<set>
#include<bitset>
#include<assert.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-6
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define NM 50005
#define nm 100005
#define pi 3.1415926535897931
using namespace std;
const int inf=1e9;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}





struct edge{int t;edge*next;}e[nm],*h[NM],*o=e;
void add(int x,int y){o->t=y;o->next=h[x];h[x]=o++;}
int n,m,_x,_y,cnt[NM],size[NM],a[11];
int smin,root,tot;
bool v[NM];
ll ans;

void dfs2(int x,int f){size[x]=1;link(x)if(!v[j->t]&&j->t!=f)dfs2(j->t,x),size[x]+=size[j->t];}
void getroot(int x,int f){
int s=tot-size[x];
link(x)if(!v[j->t]&&j->t!=f)getroot(j->t,x),s=max(s,size[j->t]);
if(s<smin)smin=s,root=x;
}

void dfs(int x,int f,int t){
inc(i,1,m)if(t<=a[i])ans+=cnt[a[i]-t];
link(x)if(!v[j->t]&&j->t!=f)dfs(j->t,x,t+1);
}
void dfs1(int x,int f,int t){
cnt[t]++;link(x)if(!v[j->t]&&j->t!=f)dfs1(j->t,x,t+1);
}

void solve(int x){
dfs2(x,0);
tot=size[x];smin=inf;
getroot(x,0);
v[root]++;cnt[0]++;
link(root)if(!v[j->t]){
dfs(j->t,root,1);
dfs1(j->t,root,1);
}
inc(i,0,tot)cnt[i]=0;
link(root)if(!v[j->t])solve(j->t);
}


int main(){
n=read();m=read();
inc(i,1,m)a[i]=read();
inc(i,2,n){_x=read();_y=read();add(_x,_y);add(_y,_x);}
solve(1);
inc(i,1,3){int t=(n-i+3)/3;printf("%.2lf\n",1.0*t*(t-1)*ans/n/(n-1));}
return 0;
}