Minimum spanning tree Kruskal hdu 5723 abandoned country

Title Link: Hdu 5723 abandoned country

The main topic: N points, M-bar, first form a minimum spanning tree, and then the minimum spanning tree to find the path length between any two points, and to expect

/************************************************************** problem:hdu 5723 user:youmi language:c++ R esult:accepted time:2932ms memory:22396k Solution: First notice that any two sides of the Benquan is not the same, so that the smallest spanning tree is unique, since the smallest spanning tree is unique,
Then the expectation is actually the only one, there is no minimum expectation. After finding the smallest spanning tree, the next question can be converted to
The average of distances between any two points is calculated on the minimum spanning tree, and for each edge, the number of times that all paths are used for this edge is
is the product of the number of points on both sides of the edge. Then the total contribution of this side is the number of times * edge right. Finally get the sum of the contributions of all sides divided by the total road
Number of diameters N (n-1)/2n∗ (n−1)/2 is the answer. Can be onon to find out. Either take a bit for root DFS and record its subtree package for each point II
The number of points included (including itself), Sum[i]sum[i], then the father side of the II points is n-sum[i]n−sum[i].
Walk the side of the statistics on the line. ****************************************************************///#pragma COMMENT (linker, "/stack:1024000000,1024000000")//#include <bits/stdc++.h>#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#include<stack>#include<Set>#include<sstream>#include<cmath>#include<queue>#include<deque>#include<string>#include<vector>#defineZeros (a) memset (A,0,sizeof (a))#defineOnes (a) memset (A,-1,sizeof (a))#defineSC (a) scanf ("%d", &a)#defineSC2 (A, b) scanf ("%d%d", &a,&b)#defineSC3 (a,b,c) scanf ("%d%d%d", &a,&b,&c)#defineSCS (a) scanf ("%s", a)#defineSclld (a) scanf ("%i64d", &a)#definePT (a) printf ("%d\n", a)#definePtlld (a) printf ("%i64d\n", a)#defineRep (i,from,to) for (int i=from;i<=to;i++)#defineIrep (I,to,from) for (int i=to;i>=from;i--)#defineMax (a) (a) > (b)? ( A):(B))#defineMin (a) < (b) ( A):(B))#defineLson (step<<1)#defineRson (lson+1)#defineEPS 1e-6#defineOO 0x3fffffff#defineTEST cout<< "*************************" <<endlConst DoublePi=4*atan (1.0);using namespaceStd;typedefLong Longll;template<classT> InlinevoidRead (T &N) {    CharCintFlag =1;  for(c = GetChar ();! (c >='0'&& C <='9'|| c = ='-'); c = GetChar ());if(c = ='-') flag =-1, n =0;Elsen = C-'0';  for(c = GetChar (); C >='0'&& C <='9'; c = GetChar ()) n = n *Ten+ C-'0'; N *=Flag;}intPow (int Base, ll N,intmo) {    if(n = =0)return 1; if(n = =1)return Base%mo; intTMP = Pow (Base, N >>1, MO); TMP= (LL) TMP * TMP%mo; if(N &1) TMP = (LL) TMP *Base%mo; returntmp;}//***************************intn,m;Const intmaxn=200000+Ten;intT,HEAD[MAXN],FA[MAXN],VIS[MAXN];structedge{intu,v;    ll W; voidInitint_u,int_v,ll _w) {u=_u,v=_v,w=_w; }}EDGE[MAXN*Ten];intGet_f (intu) {    returnu==fa[u]?u:fa[u]=Get_f (Fa[u]);}BOOLCMP (Edge A,edge b) {returna.w<B.W;}structside{intV,next; ll W;} E[MAXN*Ten];voidinit () {T=0;    Ones (head); Zeros (VIS);}voidBuildintUintv,ll W) {E[T].V=v; E[T].W=W; E[t].next=Head[u]; Head[u]=t++;}voidKruskal () {Sort (Edge+1, edge+1+m,cmp); Rep (I,0, N) fa[i]=i; ll Sum=0; Rep (I,1, M) {        intf1=Get_f (EDGE[I].U); intF2=Get_f (EDGE[I].V); if(f1!=F2) {FA[F2]=F1;            Build (EDGE[I].U,EDGE[I].V,EDGE[I].W);            Build (EDGE[I].V,EDGE[I].U,EDGE[I].W); Sum+=EDGE[I].W; //j + +;}} printf ("%i64d", sum);}DoubleTt;ll DFS (intu) {ll num=1, temp=0; Vis[u]=1;  for(intI=head[u];~i;i=E[i].next) {        intv=e[i].v; ll W=E[I].W; if(!Vis[v]) {Temp=Dfs (v); TT+=1.0*w* (1.0* (n-temp) *temp); Num+=temp; }    }    returnnum;}voidSovle () {TT=0; DFS (1); TT=tt*2.0/(1.0*n* (n1)); printf ("%.2lf\n", TT);}intMain () {#ifndef Online_judge freopen ("In.txt","R", stdin); #endif    intt_t; scanf ("%d",&t_t);  for(intKase=1; kase<=t_t;kase++) {SC (n);        SC (m);        Init (); intu,v;        ll W; Rep (I,1, M)            {SC2 (u,v); Sclld (w);        Edge[i].init (U,V,W);        } Kruskal ();    Sovle (); }}

Minimum spanning tree Kruskal hdu 5723 abandoned country