POJ3107Godfather找重心

易水人去，明月如霜。

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 aren 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 gangsterb_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

61 22 32 53 43 6

Sample Output

2 3题意：警察想抓捕黑手党老大。现在警察们认为黑手党内部是树形结构，每个人看做节点。删除一个节点，这棵树就分为几个连通分量。现在对于每个节点，删去后得到的连同分量的最大值为t。t的值最小的是哪些人，都输出。思路：求出所有的重心再排序输出代码：
#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;int read(){    char ch;int s=0,f=1;    ch=getchar();    while(ch>'9'||ch<'0'){ if(ch=='-') f*=-1; ch=getchar(); }    while(ch>='0'&&ch<='9'){ s=s*10+ch-48; ch=getchar(); }    return s*f;}struct node { int y,nxt;}e[50000*2+10];int head[50005],ans[50005],top,cnt;int size[50005],nm,mx[50005];int n;void add(int x,int y){    cnt++;    e[cnt].nxt=head[x],e[cnt].y=y;    head[x]=cnt;}void getson(int x,int fa){    size[x]=1;    for(int i=head[x];i;i=e[i].nxt)    {        int v=e[i].y;        if(v==fa) continue;        getson(v,x);        size[x]+=size[v];        mx[x]=max(mx[x],size[v]);    }    mx[x]=max(mx[x],n-size[x]);    if(mx[x]<nm){        nm=mx[x];        top=0;        ans[top++]=x;    }    else if(nm==mx[x]) ans[top++]=x;}int main(){    n=read();    for(int i=1;i<n;i++)    {        int u,v;        u=read(),v=read();        add(u,v);        add(v,u);    }    nm=100000000;    getson(1,0);   // printf("%d",top);    sort(ans,ans+top);    for(int i=0;i<top;i++)        printf("%d ",ans[i]); return 0;}