POJ1655 Balancing Act (center of gravity of the tree)

Test instructions

Give you a tree, ask for the center of the tree

If there is more than one, the output ordinal is the smallest

Ideas:

The center of the tree is the smallest number of nodes in all subtrees that are rooted in it.

Tree DP is easy to solve

/************************************************author:d evil*********************************************** * */#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#include<queue>#include<Set>#include<map>#include<string>#include<cmath>#include<stdlib.h>using namespaceStd;typedefLong LongLL;Const intinf=0x3f3f3f3f;Const intmod=1e9+7;Const intn=2e4+Ten;intn,sum[n],ans,p;BOOLVis[n];vector<int>Eg[n];voidDfsintu) {    intMa=0; Vis[u]=1;  for(intI=0; I<eg[u].size (); i++)    {        intv=Eg[u][i]; if(Vis[v])Continue;        DFS (v); Sum[u]+=sum[v]+1; Ma=max (ma,sum[v]+1); } ma=max (ma,n-sum[u]-1); if(ma<ans| | ma==ans&&u<p) {ans=Ma; P=u; }}intMain () {//freopen ("In.txt", "R", stdin);    intT,x,y; scanf ("%d",&t);  while(t--) {scanf ("%d",&N); Ans=inf;  for(intI=1; i<=n;i++) {eg[i].clear (); Sum[i]=0; Vis[i]=0; }         for(intI=1; i<n;i++) {scanf ("%d%d",&x,&y);            Eg[x].push_back (y);        Eg[y].push_back (x); } DFS (1); printf ("%d%d\n", P,ans); }    return 0;}

POJ1655 Balancing Act (center of gravity of the tree)