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=,f=; char ch=getchar();
     while (ch<'' || ch>'') {if (ch=='-') f=-; ch=getchar();}
     while (ch>='' && ch<='') {x=x*+ch-''; ch=getchar();}
     return x*f;
 }
 LL n,m,cnt=;
 double ans,tot;
 struct edge{
     int from,to,next;
     LL v,T;
 }e[];
 LL size[],head[],deep[];
 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]=;
     for (int i=head[x];i;i=e[i].next)
     {
         if (e[i].to==fa) continue;
         deep[e[i].to]=deep[x]+;
         dfs(e[i].to,x);
         size[x]+=size[e[i].to];
     }
 }
 int main(int argc, char const *argv[])
 {
     n=ra(); tot=n*(n-)*(n-)/6.0;
     for (int i=; i<n; i++)
     {
         int x=ra(),y=ra(),v=ra();
         insert(x,y,v); insert(y,x,v);
     }
     dfs(,);
     for (int i=; i<=cnt; i+=)
     {
         int now=i/,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-);
         ans+=e[i].T*e[i].v;
     }
     m=ra();
     for (int i=; i<=m; i++)
     {
         int x=ra(),y=ra();
         LL now=x*;
         ans-=(e[now].v-y)*e[now].T;
         e[now].v=y;
         printf("%.10lf\n",(double)ans/tot);
     }
     return ;
 }