2 seconds
256 megabytes
standard input
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.
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).
好题，首先我们要求在没有修改的情况下，选三个点的期望代价，这个代价来自于边，所以我们要统计出每一条边的贡献值，最后除以Ｃ（３，ｎ）就是期望值
这个可以先树形ｄｐ预处理，然后每次修改边，就把期望值减去相差的那部分，就是答案了
/*************************************************************************
> File Name: cf_d.cpp
> Author: ALex
> Mail: 405045132@qq.com
> Created Time: 2014年12月31日 星期三 09时16分18秒
************************************************************************/
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 100010;
int num[N];
int par[N];
int head[N], tot;
__int64 n;
int a[N], b[N], l[N];
double ans;
struct node
{
int weight;
int next;
int to;
}edge[N << 1];
void addedge(int from, int to,int weight)
{
edge[tot].weight = weight;
edge[tot].to = to;
edge[tot].next = head[from];
head[from] = tot++;
}
int dfs(int u, int fa)
{
int child = 1;
par[u] = fa;
for (int i = head[u]; ~i; i = edge[i].next)
{
int v = edge[i].to;
if (v == fa)
{
continue;
}
int num_v = dfs(v, u);
double cnt = (double)((n - num_v) * (num_v - 1) * num_v * 1.0 / 2);
cnt += (double)((n - num_v) * (n - num_v - 1) * num_v * 1.0 / 2);
cnt *= 2;
ans += cnt * edge[i].weight;
child += num_v;
}
num[u] = child;
// printf("node %d has %d\n", u, num[u]);
return num[u];
}
int main()
{
while (~scanf("%I64d", &n))
{
memset (num, 0, sizeof(num));
ans = 0;
tot = 0;
memset (head, -1, sizeof(head));
for (int i = 1; i <= n - 1; ++i)
{
scanf("%d%d%d", &a[i], &b[i], &l[i]);
addedge(a[i], b[i], l[i]);
addedge(b[i], a[i], l[i]);
}
dfs(1, -1);
double cnt = (n - 1) * (n - 2) * n * 1.0 / 6;
ans /= cnt;
int q;
int r, w;
scanf("%d", &q);
while (q--)
{
scanf("%d%d", &r, &w);
int dis = l[r] - w;
l[r] = w;
int son;
if (par[a[r]] == b[r])
{
son = a[r];
}
else
{
son = b[r];
}
double x = (double)(n - num[son]) * (num[son] - 1) * num[son] * 1.0/ 2;
x += (double)(n - num[son] - 1) * (n - num[son]) * num[son] *1.0 / 2;
x *= 2;
x *= dis;
x /= cnt;
ans -= x;
printf("%.10f\n", ans);
}
}
return 0;
}