URAL 1056(樹形DP)

1056. Computer Net

Time limit: 2.0 second
Memory limit: 64 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

題意：1號電腦爲根節點，給出2~n號電腦的父節點，求出用那些電腦當中心，到其餘電腦的最大距離最小。

思路：一點到所有子節點的最大距離很好求，但是我們還要知道通過父節點所能到達的最大距離，所以先處理一下，求出所有點到其子節點的最大距離邊和次大距離邊，然後再分析每個節點是否在其父節點的最大距離邊上，如果在的話，就用父節點的次大距離來更新該節點，否則用父節點的最大距離來更新。

#include<stdio.h>
#include<string.h>
const int N=10001;
int dp[N][2],vis[N],head[N],num,ans;
struct edge
{
	int st,ed,next;
}e[N*4];
void addedge(int x,int y)
{
	e[num].st=x;e[num].ed=y;e[num].next=head[x];head[x]=num++;
	e[num].st=y;e[num].ed=x;e[num].next=head[y];head[y]=num++;
}
void dfs1(int u)
{
	vis[u]=1;
	int i,v;
	for(i=head[u];i!=-1;i=e[i].next)
	{
		v=e[i].ed;
		if(vis[v]==1)continue;
		dfs1(v);
		if(dp[v][1]+1>dp[u][1])
		{
			dp[u][0]=dp[u][1];
			dp[u][1]=dp[v][1]+1;
		}
		else if(dp[v][1]+1>dp[u][0])
			dp[u][0]=dp[v][1]+1;
	}
}
void dfs2(int u)
{
	vis[u]=1;
	int i,v,temp;
	for(i=head[u];i!=-1;i=e[i].next)
	{
		v=e[i].ed;
		if(vis[v]==1)continue;		
		if(dp[u][1]==dp[v][1]+1)//在父節點的最大距離邊上
			temp=dp[u][0]+1;
		else temp=dp[u][1]+1;
		if(temp>dp[v][1])  //更新v節點的最大，次大邊
        {  
            dp[v][0]=dp[v][1];  
            dp[v][1]=temp;  
        }  
        else if(temp>dp[v][0])  
        {  
            dp[v][0]=temp;  
        }  
		if(ans>dp[v][1])
			ans=dp[v][1];
		dfs2(v);
	}
}
int main()
{
	int n,i,x;
	while(scanf("%d",&n)!=-1)
	{
		memset(head,-1,sizeof(head));
		memset(dp,0,sizeof(dp));
		num=0;
		for(i=2;i<=n;i++)
		{
			scanf("%d",&x);
			addedge(i,x);
		}
		ans=99999999;
		memset(vis,0,sizeof(vis));
		dfs1(1);
		memset(vis,0,sizeof(vis));
		dfs2(1);
		if(ans>dp[1][1])ans=dp[1][1];
		for(i=1;i<=n;i++)
		{
			if(dp[i][1]==ans)
				printf("%d ",i);
		}
		printf("\n");
	}
	return 0;
}