URAL 1056 Computer Net (Shortest way)

Computer NetTime limit:2.0 Second
Memory limit:64 mbbackgroundcomputer Net is created by consecutive computer plug-up to one it has already been connecte D to the net. Each new computer gets a ordinal number, but the protocol contains the number of their parent computer in the net. Thus, protocol consists of several numbers; The first of them is all 1, because the second computer can only be connected to the first one, and the second number is 1 or 2 and so forth. The total quantity of numbers in the protocol is N? 1 ( Nis a total number of computers). For instance, protocol 1, 1, 2, 2 corresponds to the following net:

1-2-5| | 3 4

The distance between the computers is the quantity of mutual connections (between each other) in chain. Thus, in example mentioned above the distance between computers #4 and #5 are 2, and between #3 and #5 is 3. Definition.Let the center of the net being the computer which have a minimal distance to the most remote computer. In the shown example computers #1 and #2 is the centers of the net. Problemyour task is to find all the centers using the set protocol. Inputthe first line of input contains an integer N, the quantity of computers (2≤ N≤10000). Successive N? 1 lines contain protocol. Outputoutput should contain ordinal numbers of the determined net centers in ascending order. Sample

input	Output
51122	1 2

problem Source:Rybinsk State Avia Academy "Analysis" gives you a tree, n nodes, N-1 edge, next n-1 line, I th line represents the node number connected to I, you need to find a few points, so that these points from the farthest point distance from the smallest. It is the shortest way to find the diameter of the tree, record the length of the diameter, and then find the midpoint along the diameter.

#include <iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<cmath>#include<string>#include<stack>#include<queue>#include<vector>#defineINF 0x3f3f3f3f#defineMet (b) memset (a,b,sizeof a)typedefLong Longll;using namespacestd;Const intN =10005;Const intM =24005;intVis[n],dis[n],pre[n],head[n];intn,m,tot=0, SON,MAXN;structedg{intTo,next;} edg[n*N];voidAddintUintv) {edg[tot].to=v;edg[tot].next=head[u];head[u]=tot++;}voidBFsints) {met (Vis,0Met (Dis,inf), met (Pre,0); Dis[s]=0; Vis[s]=1; maxn=0; Queue<int>Q;q.push (s);  while(!Q.empty ()) {        intU=q.front (); Q.pop (); vis[u]=0;  for(inti=head[u];i!=-1; i=Edg[i].next) {            intv=edg[i].to; if(dis[v]>dis[u]+1) {Dis[v]=dis[u]+1; Maxn=max (Maxn,dis[v]);p re[v]=u; if(!Vis[v]) {Q.push (v); Vis[v]=1; }}} son=u; }}intMain () {Met (head,-1); intA[n],cnt=0; scanf ("%d",&N);  for(intI=2; i<=n;i++) {scanf ("%d",&m); add (i,m); add (m,i); } BFS (1);ints=Son; BFS (son);intT=son;//printf ("!! %d%d\n ", s,t);    if(maxn&1){        intx=maxn/2;  while(t) {if(dis[t]==x| | dis[t]==x+1) a[cnt++]=T; T=Pre[t]; }    }    Else {        intx=maxn/2;  while(t) {if(dis[t]==x) a[cnt++]=T; T=Pre[t]; }} sort (A,a+CNT);  for(intI=0; i<cnt;i++) printf ("%d", A[i]); printf ("\ n"); return 0;}

URAL 1056 Computer Net (Shortest way)