Godfather

Description

Last years Chicago was full of gangster fights and strange murders. The chief of the police got really tired of all these crimes, and decided to arrest the mafia leaders.

Unfortunately, the structure of Chicago mafia is rather complicated. There are n persons known to be related to mafia. The police have traced their activity for some time, and know that some of them are communicating with each other. Based on the data collected, the chief of the police suggests that the mafia hierarchy can be represented as a tree. The head of the mafia, Godfather, is the root of the tree, and if some person is represented by a node in the tree, its direct subordinates are represented by the children of that node. For the purpose of conspiracy the gangsters only communicate with their direct subordinates and their direct master.

Unfortunately, though the police know gangsters’ communications, they do not know who is a master in any pair of communicating persons. Thus they only have an undirected tree of communications, and do not know who Godfather is.

Based on the idea that Godfather wants to have the most possible control over mafia, the chief of the police has made a suggestion that Godfather is such a person that after deleting it from the communications tree the size of the largest remaining connected component is as small as possible. Help the police to find all potential Godfathers and they will arrest them.

Input

The first line of the input file contains n — the number of persons suspected to belong to mafia (2 ≤ n ≤ 50 000). Let them be numbered from 1 to n.

The following n − 1 lines contain two integer numbers each. The pair a_i, b_i means that the gangster a_i has communicated with the gangster b_i. It is guaranteed that the gangsters’ communications form a tree.

Output

Print the numbers of all persons that are suspected to be Godfather. The numbers must be printed in the increasing order, separated by spaces.

Sample Input

6
1 2
2 3
2 5
3 4
3 6

Sample Output

2 3

分析

題目的意思很明確，就是求所有樹的重心（再按字典序輸出）。

那麽我們先介紹一下樹的重心。
樹的重心定義為：
樹中的一個點,刪掉該點，使剩下的樹所構成的森林中最大的子樹節點數最少。
樹的重心推論：
1.設樹上的一個點S，樹上其余所有點到S點的距離之和最小，那麽S就是重心。
2.樹的重心不唯一。
那麽我們依靠定義來求樹的重心好了。
首先我們確定一個根，進行一遍dfs，回溯的時候可以遞歸統計該點不同子樹所擁有的點的數量，
然後再用（總節點數）減去（1）減去（子樹），就是其父親那邊的那棵樹的點的數量，
取min，最後求出即可。

code

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 
 5 using namespace std;
 6 
 7 const int MAXN = 50010;
 8 const int MAXM = 100010;
 9 
10 struct Edge{
11     int to,nxt;
12 }e[MAXM];
13 int head[MAXM],tot;
14 int son[MAXN];
15 int ans[MAXN],p,Ans = 1e9,n;
16 
17 inline int read() {
18     int x = 0,f = 1;char ch = getchar();
19     for (; ch<‘0‘||ch>‘9‘; ch = getchar())
20         if (ch==‘-‘) f = -1;
21     for (; ch>=‘0‘&&ch<=‘9‘; ch = getchar())
22         x = x*10+ch-‘0‘;
23     return x*f;
24 }
25 inline void init() {
26     memset(head,0,sizeof(head));
27     memset(son,0,sizeof(son));
28     tot = 0;
29 }
30 inline void add_edge(int u,int v) {
31     e[++tot].to = v,e[tot].nxt = head[u],head[u] = tot; 
32 }
33 void dfs(int u,int fa) {
34     int cnt = 0;
35     for (int i=head[u]; i; i=e[i].nxt) {
36         int v = e[i].to;
37         if (v==fa) continue;
38         dfs(v,u);
39         son[u] += son[v]+1;
40         cnt = max(cnt,son[v]+1);
41     }
42     cnt = max(cnt,n-son[u]-1);
43     if (cnt<Ans) {Ans = cnt,p = 0,ans[++p] = u;}
44     else if (cnt==Ans) {ans[++p] = u;}
45 }
46 int main() {
47     
48     while (scanf("%d",&n)!=EOF) {
49         init();
50         for (int u,v,i=1; i<n; ++i) {
51             u = read(),v = read();
52             add_edge(u,v),add_edge(v,u);
53         }
54         dfs(1,0);
55         sort(ans+1,ans+p+1);
56         for (int i=1; i<=p; ++i) 
57             printf("%d ",ans[i]);    
58         printf("\n");    
59     }    
60     return 0;
61 }

poj3107Godfather(樹的重心)