【Goodbye2014】Codeforces 500D New Year Santa Network【套路思维题】

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.

Examples

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 to
d(1, 2) + d(2, 3) + d(3, 1).

题目大意：

给你一颗树，其中有N个点，N-1条边。

每条边都有一个权值（长度），然后有Q个查询，表示编号为x的边长度变成了y.

每一次查询询问改变长度之后的图，期望花费是多少。

我们需要在这个图中每次选择三个点，然后这三个点需要花费的价值是D（a，b）+D（b，c）+D（c，a）；D（a，b）表示a和b之间的距离.

思路：

1、这种题很套路，直接考虑一条边的贡献度即可（下图样例）。



假设这条边，权值为w，那么其贡献度为：

val=w*（C(x,2)*C(y,1)+C(x,1)*C(y,2)）;

那么其对期望的贡献度为：

val/C(n,3);

2、那么我们Dfs处理出来每一条边的贡献度，并且记录每条边的单位贡献度，那么在查询的时候，直接按照倍数减少即可。

3、注意数据范围，该double的一定要double.

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
#define ll __int64
struct node
{
int from;
int to;
int w;
int num;
int next;
}e[300050];
int head[100050];
int len[100050];
double del[100050];
int cont,n;
double ans;
void add(int from,int to,int w,int num)
{
e[cont].w=w;
e[cont].to=to;
e[cont].num=num;
e[cont].from=from;
e[cont].next=head[from];
head[from]=cont++;
}
double C(int n,int m)
{
if(m>n)return 0;
if(m==0)return 1;
if(m==1)
{
return (double)n;
}
if(m==2)
{
return (double)n*(double)(n-1)/double(2*1.0);
}
if(m==3)
{
return (double)n*(double)(n-1)*(double)(n-2)/(double)(3*2*1.0);
}
}
ll Dfs(int u,int from)
{
ll cnt=1;
ll tmp;
for(int i=head[u];i!=-1;i=e[i].next)
{
int v=e[i].to;
int w=e[i].w;
if(v==from)continue;
else
{
tmp=Dfs(v,u);
double fenzi=(double)w*((double)C(tmp,2)*(double)C(n-tmp,1)+(double)C(tmp,1)*(double)C(n-tmp,2));
double fenmu=(double)C(n,3);
ans+=fenzi*2.0/fenmu*1.0;
del[e[i].num]=fenzi*2.0/(fenmu*1.0*(double)w);
cnt+=tmp;
}
}
return cnt;
}
int main()
{
while(~scanf("%d",&n))
{
cont=0;
memset(head,-1,sizeof(head));
ans=0;
for(int i=0;i<n-1;i++)
{
int x,y,w;
scanf("%d%d%d",&x,&y,&w);
add(x,y,w,i);
add(y,x,w,i);
len[i]=w;
}
Dfs(1,-1);
int q;
scanf("%d",&q);
while(q--)
{
int x,y;
scanf("%d%d",&x,&y);
x--;
ans-=(len[x]-y)*del[x];
len[x]=y;
printf("%lf\n",ans);
}
}
}