2 seconds
256 megabytes
standard input
standard output
New Year is coming in Tree World! In this world, as the name implies, there are
n cities connected by
n - 1 roads, and for any two distinct cities there always exists a path between them. The cities are numbered by integers from 1 to
n, and the roads are numbered by integers from
1 to n - 1. Let's define
d(u, v) as total length of roads on the path between city
u and city v.
As an annual event, people in Tree World repairs exactly one road per year. As a result, the length of one road decreases. It is already known that in the
i-th year, the length of the
ri-th road is going to become
wi, which is shorter than its length before. Assume that the current year is year
1.
Three Santas are planning to give presents annually to all the children in Tree World. In order to do that, they need some preparation, so they are going to choose three distinct cities
c1,
c2, c3 and make exactly one warehouse in each city. The
k-th (1 ≤ k ≤ 3) Santa will take charge of the warehouse in city
ck.
It is really boring for the three Santas to keep a warehouse alone. So, they decided to build an only-for-Santa network! The cost needed to build this network equals to
d(c1, c2) + d(c2, c3) + d(c3, c1)
dollars. Santas are too busy to find the best place, so they decided to choose
c1, c2, c3 randomly uniformly over all triples of distinct numbers from
1 to n. Santas would like to know the expected value of the cost needed to build the network.
However, as mentioned, each year, the length of exactly one road decreases. So, the Santas want to calculate the expected after each length change. Help them to calculate the value.
The first line contains an integer n (3 ≤ n ≤ 105) — the number of cities in Tree World.
Next n - 1 lines describe the roads. The
i-th line of them (1 ≤ i ≤ n - 1) contains three space-separated integers
ai,
bi,
li (1 ≤ ai, bi ≤ n,
ai ≠ bi,
1 ≤ li ≤ 103), denoting that the
i-th road connects cities
ai and
bi, and the length of
i-th road is li.
The next line contains an integer q (1 ≤ q ≤ 105) — the number of road length changes.
Next q lines describe the length changes. The
j-th line of them (1 ≤ j ≤ q) contains two space-separated integers
rj,
wj (1 ≤ rj ≤ n - 1,
1 ≤ wj ≤ 103). It means that in the
j-th repair, the length of the
rj-th road becomes
wj. It is guaranteed that
wj is smaller than the current length of the
rj-th road. The same road can be repaired several times.
Output q numbers. For each given change, print a line containing the expected cost needed to build the network in Tree World. The answer will be considered correct if its absolute and relative error doesn't exceed
10 - 6.
3 2 3 5 1 3 3 5 1 4 2 2 1 2 2 1 1 1
14.0000000000 12.0000000000 8.0000000000 6.0000000000 4.0000000000
6 1 5 3 5 3 2 6 1 7 1 4 4 5 2 3 5 1 2 2 1 3 5 4 1 5 2
19.6000000000 18.6000000000 16.6000000000 13.6000000000 12.6000000000
Consider the first sample. There are 6 triples: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1). Because
n = 3, the cost needed to build the network is always
d(1, 2) + d(2, 3) + d(3, 1) for all the triples. So, the expected cost equals to
d(1, 2) + d(2, 3) + d(3, 1).
可以知道，如果是完全图，对于每条边，会经过它n-2次，所以，在原图中，dfs出每条边去掉后，分成的两部分各有多少个点，假设各有a,b个点，那么这条边将会被经过
ab(n-2)次
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
using namespace std;
struct Edge
{
int to,id;
Edge(int to,int id)
{
this->to=to;
this->id=id;
}
};
vector<Edge>edge[100010];
int cnt[100010];
int dfs(int pre,int from)
{
int sum=0;
for(int i=0;i<edge[from].size();i++)
{
int to=edge[from][i].to,id=edge[from][i].id;
if(to==pre)
continue;
sum+=cnt[id]=dfs(from,to);
}
return sum+1;
}
struct Road
{
int a,b,w;
friend istream & operator >>(istream &is,Road &x)
{
is>>x.a>>x.b>>x.w;
return is;
}
}road[100010];
int main()
{
int n;
cin>>n;
for(int i=1;i<n;i++)
{
cin>>road[i];
edge[road[i].a].push_back(Edge(road[i].b,i));
edge[road[i].b].push_back(Edge(road[i].a,i));
}
dfs(0,1);
double cn3=double(n)*(n-1)*(n-2)/6,ans=0;
for(int i=1;i<=n;i++)
ans+=(double)cnt[i]*(n-cnt[i])*(n-2)*road[i].w;
ans/=cn3;
int q;
cin>>q;
while(q--)
{
int id,w;
cin>>id>>w;
ans+=(double)cnt[id]*(n-cnt[id])*(n-2)*(w-road[id].w)/cn3;
road[id].w=w;
cout<<fixed<<setprecision(9)<<ans<<endl;
}
}