Good Bye 2014 D. New Year Santa Network 树形dp

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

3
2 3 5
1 3 3
5
1 4
2 2
1 2
2 1
1 1

output

14.0000000000
12.0000000000
8.0000000000
6.0000000000
4.0000000000

input

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

output

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 tod(1, 2) + d(2, 3) + d(3, 1).

题意，给出一棵树，每边条有对应的边树，每次改变一个边的权值，要求随便选三个点，求三个点之间距离权值合的期望。

因为树的结构没有变化，所以可以先用树形dp预处理出每条边对最终结果影响，也就是每条边最终会算多少次，这样可以每次查询的时候只用o(1)复杂度就可以了。

怎么求每条边的个数呢，对任一条连，其左端点的连通块结点个数为x,则其右端点连通块个数为n - x,设为y,则，只要左边选两个点，右边选一个点，或左连选两个点，右边先一个点，这样就有c(x,2) * y + c(y,2) * x个，由于，三角形有两个边，所以还要* 2,化简可以得到sum( x *  (n - x) * (n - 2)};最终，由有n个点，任选3个点，有C(n,3)个三角形，所以还要除以C(n,3),约去(n - 2) ,得到sum{6 * x * (n-x) }/ (n * (n -1));用树形dp,只要o(n
+q)的复杂度就可以了。

#define N 100005
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n,a,b,c,cnt
,q;
ll num
,e
,ans,mo;
vector<pii> p
;
bool vis
;
void DFS(int x){
vis[x] = true;
cnt[x] = 1;
FI(p[x].size()){
int t = p[x][i].first;
if(!vis[t]){
DFS(t);
num[p[x][i].second] = 6ll * (ll)cnt[t] * ((ll)n - (ll)cnt[t]);
cnt[x] += cnt[t];
}
}
}
int main()
{
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
while(S(n)!=EOF)
{
For(i,1,n) p[i].clear();
For(i,1,n - 1){
S2(a,b);S(c);
e[i] = (ll)c;
p[a].push_back(mp(b,i));
p[b].push_back(mp(a,i));
}
fill(vis,false);
DFS(1);
ans = 0;
For(i,1,n-1) {
ans += num[i] * e[i];
}
S(q);
double factor = 1.0/n/(n-1);
while(q--){
S2(a,b);
ans -= (e[a] - (ll)b) * num[a];
e[a] = b;
printf("%.6f\n",(double)ans * factor);
}
}
//fclose(stdin);
//fclose(stdout);
return 0;
}