树的中心

Lshain — Mon, 20 Jun 2011 09:30:00 GMT

http://acm.timus.ru/problem.aspx?space=1&num=1056

1056. Computer Net

Time Limit: 2.0 second
Memory Limit: 16 MB

Background

Computer net is created by consecutive computer plug-up to one that has already been connected to the net. Each new computer gets an ordinal number, but the protocol contains the number of its parent computer in the net. Thus, protocol consists of several numbers; the first of them is always 1, because the second computer can only be connected to the first one, the second number is 1 or 2 and so forth. The total quantity of numbers in the protocol is N − 1 (N is 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 is 2, and between #3 and #5 is 3.

Definition. Let the center of the net be the computer which has a minimal distance to the most remote computer. In the shown example computers #1 and #2 are the centers of the net.

Problem

Your task is to find all the centers using the set protocol.

Input

The first line of input contains an integer N, the quantity of computers (2 ≤ N ≤ 10000). Successive N − 1 lines contain protocol.

Output

Output should contain ordinal numbers of the determined net centers in ascending order.

Sample

input	output
5 1 1 2 2	1 2

给定一棵树，求该树的中心。
中心定义：到其它点的距离最大值最小
树的中心即为树中最长链的中心，2次dfs即可

Code
#include
struct Node{
   int n;
   int next;
}EE[20001];

int ind;
int n;
int top;
int head[10001];
int check[10001];
int pre[10001];
int maxDep;
void dfs(int u, int t){
   int q;
   check[u] = 1;
   if(t > maxDep){
      maxDep = t;
      ind = u;
   }
   q = head[u];
   while(q != -1){
      if(!check[EE[q].n]){
         pre[EE[q].n] = u;
         dfs(EE[q].n, t + 1);
      }
      q = EE[q].next;
   }
}

int main(int argc, char* argv[], char* env[])
{
   int i = 0;
   int j = 0;
   while(scanf("%d", &n) != EOF){
      for(i = 1; i <= n; i++){
         head[i] = -1;
         check[i] = 0;
      }
      top = 0;
      for(i = 2; i <= n; i++){
         scanf("%d", &j);
         EE[top].n = i;
         EE[top].next = head[j];
         head[j] = top++;
         EE[top].n = j;
         EE[top].next = head[i];
         head[i] = top++;
      }
      maxDep = -1;
      dfs(1, 1);
      for(i = 1; i <= n; i++){
         check[i] = 0;
         pre[i] = -1;
      }
      maxDep = -1;
      dfs(ind, 1);
      i = ind;
      top = 0;
      while(i != -1){
         check[top++] = i;
         i = pre[i];
      }
      if(top&1){
         printf("%d\n", check[top>>1]);
      }
      else{
         printf("%d %d\n", check[(top>>1) - 1], check[top>>1]);
      }
   }
   return 0;
}

Lshain 2011-06-20 17:30 发表评论