bupt spring training for 2015 #2nd J
给两棵树,分别有 n,m 个节点(1 ≤ N, Q ≤ 4 × 10^4),等概率连接属于不同树的两个节点,求新树的直径(最远两点的距离)的期望。
#include<bits/stdc++.h>
#define ll long long
#define N 100005
using namespace std;
struct edge{int to,next;}e[N<<1];
int head[N],cnt;
void add(int u,int v){e[cnt]=(edge){v,head[u]};head[u]=cnt++;}
void init(){memset(head,-1,sizeof head);cnt=0;}
int d[N],d2[N],ds[N];
int deep,dex;
ll s[N],ans;
void dfs(int x,int fa){
if(d[x]>=deep)deep=d[x],dex=x;
for(int i=head[x];~i;i=e[i].next){
int v=e[i].to;
if(v==fa)continue;
d[v]=d[x]+1;
dfs(v,x);
}
}
int getDep(int x,int ed){
deep=0;d[x]=0;dfs(x,0);
d[dex]=0;dfs(dex,0);
for(int i=x;i<=ed;++i)d2[i]=d[i];
d[dex]=0;dfs(dex,0);
for(int i=x;i<=ed;++i)ds[i]=max(d[i],d2[i]);
return deep;
}
int main(){
int n,m;
while(~scanf("%d%d",&n,&m)){
init();ans=0;
for(int u,v,i=1;i<n;++i)scanf("%d%d",&u,&v),add(u,v),add(v,u);
for(int u,v,i=1;i<m;++i)scanf("%d%d",&u,&v),add(u+n,v+n),add(v+n,u+n);
int len=max(getDep(1,n),getDep(n+1,n+m));
sort(ds+1,ds+1+n);
for(int i=1;i<=n;++i)s[i]=s[i-1]+ds[i];
for(int i=1;i<=m;++i){
int b=lower_bound(ds+1,ds+1+n,len-ds[i+n]-1)-ds-1;
ans+=(ll)b*len+(ll)(n-b)*ds[i+n]+(n-b)+s[n]-s[b];
}
printf("%.3f\n",ans*1./(n*m));
}
return 0;
}