# Codeforces E. famil Door and Roads expectations

Source: Internet
Author: User

A tree, n nodes, the edge length is 1, there are Q queries, each query given u,v (u! = V), ask in the tree the probability plus an edge, if the u,v in a ring, then this method of addition is legal, the value is the length of the ring, all the legal edge of the occurrence of equal probability, ask the value of expectations.

2 <= n,m <= 10^5

For u,v the original path on the edge must be within the ring, the contribution of 1, the new addition of the edge must also be within the ring, contribute to 1, the rest of the side of the contribution of the line

Consider 2 scenarios:

1.LCA (u,v) Not equal to U and V

2.lca (u,v) for u or V

Code:

`//File Name:cf629E.cpp//Created time:2017 January 05 Thursday 17:20 29 seconds#include<bits/stdc++.h>#defineLL Long Longusing namespacestd;Const intMAXN =100000+2;intSIZ[MAXN],DIS[MAXN];intpa[maxn][ -]; LL F[maxn],g[maxn];vector<int>G[MAXN];voidDfs0 (intUintp) {Siz[u]=1; Dis[u]= Dis[p] +1; F[u]=0;  for(intI=0; I<g[u].size (); + +i) {        intv =G[u][i]; if(v = = p)Continue; pa[v][0] =u;        Dfs0 (V,u); Siz[u]+=Siz[v]; F[u]+ = F[v] +Siz[v]; }}voidDFS1 (intUintPintN) {     for(intI=0; I<g[u].size (); + +i) {        intv =G[u][i]; if(v = = p)Continue; G[V]= n + g[u]-2*Siz[v];    DFS1 (V,u,n); }}voidCAL_PA (intN) {     for(intj=1;(1&LT;&LT;J) <=n;++K) {         for(intI=1; i<=n;++i) {            if(pa[i][j-1] != -1) Pa[i][j]= pa[pa[i][j-1]][j-1]; }    }}intCal_lca (intAintb) {    if(Dis[a] <dis[b]) swap (A, b); intCNT =0;  for(;(1&LT;&LT;CNT) <=dis[a];++CNT); --CNT;  for(intj=cnt;j>=0;--j) {        if(Dis[a]-(1&LT;&LT;J) >=Dis[b]) a=Pa[a][j]; }    if(A = = b)returnA;  for(intj=cnt;j>=0;--j) {        if(Pa[a][j]! =-1&& Pa[a][j]! =Pa[b][j]) a= Pa[a][j],b =Pa[b][j]; }    returnpa[a][0];}intCalintUintv) {    intCNT =0;  for(;(1&LT;&LT;CNT) <=dis[u];++CNT); --CNT;  for(intj=cnt;j>=0;--j) {        if(Dis[u]-(1&LT;&LT;J) >dis[v]) u=Pa[u][j]; }    returnu;}voidSolveintNintm) {memset (PA,-1,sizeof(PA)); Dfs0 (1,0);//DIS,SIZ,IN,F,PA[I][0],DEPg[1] = f[1]; DFS1 (1,0, n);//gCAL_PA (n);//PA    intu,v,lca,w;  while(m--) {scanf ("%d%d",&u,&v); LCA=Cal_lca (U,V); if(LCA! = u && LCA! =v) {            intTMP = Dis[u] + dis[v]-2* DIS[LCA] +1; DoubleAns = tmp + (F[u] +0.0)/Siz[u] + (F[v] +0.0) /Siz[v]; printf ("%.15f\n", ans); }        Else{            if(LCA = =u) swap (u,v); W=cal (U,V); if(pa[w][0] !=v) {printf ("-1"); return ; }            intTMP = Dis[u]-DIS[V] +1; DoubleAns = tmp + (F[u] +0.0)/Siz[u] + (G[v]-f[w]-siz[w] +0.0)/(N-siz[w]); printf ("%.15f\n", ans); }    }}intMain () {intn,m; scanf ("%d%d",&n,&m);  for(intI=1, u,v;i<n;++i) {scanf ("%d%d",&u,&v);        G[u].push_back (v);    G[v].push_back (U);    } solve (n,m); return 0;}`

Codeforces E. famil Door and Roads expectations

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