cf 500 D. New Year Santa Network

直接按边分，2个点在边的一边，1个在另一边，组合出来就是这个边对答案的贡献，权值换了就再重新算个数而已。

#include <bits/stdc++.h>
#define LL long long
using namespace std;
inline int ra()
{
    int 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=1;
double ans,tot;
struct edge{
    int from,to,next;
    LL v,T;
}e[200005];
LL size[100005],head[100005],deep[100005];
void insert(int x, int y, int v)
{
    e[++cnt].next=head[x]; e[cnt].from=x; e[cnt].to=y; e[cnt].v=v; head[x]=cnt;
}
void dfs(int x, int fa)
{
    size[x]=1;
    for (int i=head[x];i;i=e[i].next)
    {
        if (e[i].to==fa) continue;
        deep[e[i].to]=deep[x]+1;
        dfs(e[i].to,x);
        size[x]+=size[e[i].to];
    }
}
int main(int argc, char const *argv[])
{
    n=ra(); tot=n*(n-1)*(n-2)/6.0;
    for (int i=1; i<n; i++)
    {
        int x=ra(),y=ra(),v=ra();
        insert(x,y,v); insert(y,x,v);
    }
    dfs(1,0);
    for (int i=2; i<=cnt; i+=2)
    {
        int now=i/2,x=e[i].from,y=e[i].to;
        if (deep[x]>deep[y]) swap(x,y);
        LL s1=n-size[y],s2=size[y];
        e[i].T+=s1*s2*((LL)n-2);
        ans+=e[i].T*e[i].v;
    }
    m=ra();
    for (int i=1; i<=m; i++)
    {
        int x=ra(),y=ra();
        LL now=x*2;
        ans-=(e[now].v-y)*e[now].T;
        e[now].v=y;
        printf("%.10lf
",(double)ans/tot);
    }
    return 0;
}