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 citiesc1, 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).
题意是在一棵树上随机取三点c1,c2,c3，计算dist(c1,c2)+dist(c1,c3)+dist(c2,c3)的期望。还有带边权修改的，每一个修改输出一个答案
一开始看懂题意我都吓傻了……不过很快意识到大概是有什么结论
然后开始随便乱画……最后发现把这三条路径描出来，每条边都恰好被经过两次。这个理论证明我不会诶，不过事实证明这是对的
再考虑怎么统计答案
树的形态是不变的，所以可以直接搞出所有方案中每条边被经过的次数，并且这个数字是不会变的
考虑一条边到底在统计的时候被统计几次：显然路径有经过这条边，那么边的两端的两块联通块都有至少一个点。只可能是一边一个一边两个了
然后排列组合随便搞搞就好了
修改就更简单了，直接在答案里减去(原来的权值-修改的值)*统计次数，这个O(1)就完了
非常蛋疼的是这题要爆double……printf我又不会输出……cout保留小数搞得我蛋都疼了
最后还是wa……把int全替换成LL就A了……
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<set>
#include<map>
#include<ctime>
#include<iomanip>
#define LL long long
#define inf 0x7ffffff
#define N 1000010
using namespace std;
inline LL read()
{
LL x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
LL n,m,cnt;
long double ans,todel;
struct edge{
LL from,to,next,v;
LL rep;
}e[4*N];
LL son[N],head[N],dep[N];
bool mrk[N];
inline void ins(LL u,LL v,LL w)
{
e[++cnt].to=v;
e[cnt].from=u;
e[cnt].v=w;
e[cnt].next=head[u];
head[u]=cnt;
}
inline void insert(LL u,LL v,LL w)
{
ins(u,v,w);
ins(v,u,w);
}
inline void dfs(LL x,LL d)
{
if (mrk[x])return;
mrk[x]=1;son[x]=1;dep[x]=d;
for (LL i=head[x];i;i=e[i].next)
if (!mrk[e[i].to])
{
dfs(e[i].to,d+1);
son[x]+=son[e[i].to];
}
}
int main()
{
n=read();todel=(long double)n*(n-1)*(n-2)/6.0;
for (LL i=1;i<n;i++)
{
LL x=read(),y=read(),z=read();
insert(x,y,z);
}
dfs(1,1);
for (LL i=2;i<=cnt;i+=2)
{
LL now=i/2,x=e[i].from,y=e[i].to;
if (dep[x]>dep[y])swap(x,y);
LL s1=n-son[y],s2=son[y];
e[i].rep+=(long double)s1*s2*(s2-1)+s2*s1*(s1-1);
ans+=(long double)e[i].rep*e[i].v;
}
m=read();
cout<<setiosflags(ios::fixed)<<setprecision(10);
for (LL i=1;i<=m;i++)
{
LL x=read(),y=read();
LL now=x*2;
ans-=(long double)(e[now].v-y)*e[now].rep;
e[now].v=y;
cout<<ans/todel<<endl;
}
return 0;
}
