D. New Year Santa Network

原题

Time Limit 2s

Memory Limit 256M

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.

Examples

input1

output1

14.0000000000
12.0000000000
8.0000000000
6.0000000000
4.0000000000

input2

output2

19.6000000000
18.6000000000
16.6000000000
13.6000000000
12.6000000000

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

有一棵有边权的树，每次修改一条边权的长度，并输出这时任意三点 $a,b,c$ 之间的期望距离， $d(a,b)+d(a,c)+d(b,c)$ ，也就是平常所说的平均值

Solution

任意3个点对于一条边，有两种情况，即经过2次，或不经过

经过两次，当然还有考虑两个点在左在右的问题

2edges

不经过，不需要考虑

0edge

那么我们只需枚举每条边被使用的次数乘以改变的权值，最终除以点对 $(a,b,c)$ 的个数 $\frac{n*(n-1)*(n-2)}{6}$ 即可

Code

#include <cstdio>
#include <vector>
#include <iostream>
#include <iomanip>
#define N 100005
#define LL long long

using namespace std;

int n, q, a[N], b[N], l[N], size[N], si, o, len;
double ans;
vector <int> e[N];

void dfs(int x, int fa) {
	vector <int> :: iterator it;
	size[x] = 1;
	for(it = e[x].begin(); it != e[x].end(); it++) {
		int t = *it;
		if (t == fa) continue;
		dfs(t, x);
		size[x] += size[t];
	}
}

int main(){
	scanf("%d", &n);
	for (int i = 1; i < n; ++i) {
		scanf("%d%d%d", &a[i], &b[i], &l[i]);
		e[a[i]].push_back(b[i]);
		e[b[i]].push_back(a[i]);
	}
	dfs(1, 1);
	LL psc = (LL)n * (LL)(n - 1) * (LL)(n - 2) / 6;//点对数量
	for (int i = 1; i < n; ++i) {
		if (size[a[i]] < size[b[i]]) si = size[a[i]];
		else si = size[b[i]];
		ans += (double)l[i] * (double)si * (double)(n - si) * (double)(n - si - 1) / psc;//两个点在上方，一个点在下方
		ans += (double)l[i] * (double)si * (double)(si - 1) * (double)(n - si) / psc;//两个点在下方，一个点在上方
	}
	scanf("%d", &q);
	while (q--) {
		scanf("%d%d", &o, &len);
		if (size[a[o]] < size[b[o]]) si = size[a[o]];
		else si = size[b[o]];
		ans += (double)(len - l[o]) * (double)si * (double)(n - si) * (double)(n - si - 1) / psc;//更新
		ans += (double)(len - l[o]) * (double)si * (double)(si - 1) * (double)(n - si) / psc;//更新
		l[o] = len;
		cout.setf(ios::fixed);
		cout << fixed << setprecision(7) << (double) ans << endl;//CF上居然不能用printf("%.7Lf\n", ans);
	}
	return 0;
}

CodeForces 500D

D. New Year Santa Network

原题

题意

Solution

Code

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