CF~Good Bye 2014 D. New Year Santa Network

D. New Year Santa Network

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

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.

Input

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

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.

Sample test(s)

input

32 3 51 3 351 42 21 22 11 1

output

14.000000000012.00000000008.00000000006.00000000004.0000000000

input

61 5 35 3 26 1 71 4 45 2 351 22 13 54 15 2

output

19.600000000018.600000000016.600000000013.600000000012.6000000000

Note

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).

       题意：有n各点，靠n-1条边全联通，第i条边有权值Wi；从中挑三个点a、b、c问dis(a,b)+dis(b,c)+dis(a,c)期望值；有q次修路，使得第i条权值改变，求q次改变后所求期望值。

       题解：对于任意两点(a,b)可与余下(n-2)点构成可能情况，即每对(a,b)求和(n-2)次; 期望E=[(n-2)*sum_length(a,b) ]/[n*(n-1)*(n-2)/(3*2*1)]=6.0*sum_length(a,b)/n/(n-1) ;        

       对于第i边连接的两点(u,v)，设以u、v为根的子树分别有tree_u、tree_v个点，则有：第i条边在sum_length出项次数为tree_u*tree_v；易知 tree_v=(n-treeu) 。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>#include <cctype>#include <queue>#include <stack>#include <vector>#define INF 0x7fffffff#define maxn 987654321#define sizeN 300009#define eps (1e-9)#define clearto(s,x) memset(s,x,sizeof(s))using namespace std;typedef long long llong;int n,m,tot;int length[sizeN]={0};llong used[sizeN]={0};typedef pair<int,int> node;vector<node> vec[sizeN];int dfs(int fa,int lst,int id){    int szA=1;    int chl,loc;    for(int i=0;i<vec[fa].size();i++)    {        chl=vec[fa][i].first;        loc=vec[fa][i].second;        if(chl==lst)    continue;        szA+=dfs(chl,fa,loc);    }    used[id]=(llong)szA*(n-szA)*2;    // 若这里不*2，对应下面tmp系数变为6.0；也会WA，醉了    return szA;}int main(){//freopen("E:\DATA.txt","r",stdin);int i,j;int a,b,l;scanf("%d",&n);for(i=1;i<n;i++){    scanf("%d %d %d",&a,&b,&l);    vec[a].push_back(make_pair(b,i));    vec[b].push_back(make_pair(a,i));    length[i]=l;}dfs(1,-1,0);    llong sum=0;    for(i=1;i<n;i++){        sum=sum+length[i]*used[i];        //cout<<sum<<' '<<length[i]<<' '<<used[i]<<endl;    }    double tmp= 3.0/n/(n-1);            //若写成3.0/(n*(n-1))是要挂的哦    scanf("%d",&tot);    while(tot--)    {        scanf("%d %d",&a,&b);        sum=sum-(used[a]*(length[a]-b));        printf("%.8lf\n",(double)sum*tmp);        length[a]=b;    }return 0;}