[POJ] 3107 Godfather

Godfather
Time Limit: 2000MS      Memory Limit: 65536K
Total Submissions: 8007     Accepted: 2825
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 ai, bi means that the gangster ai has communicated with the gangster bi. 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
Source

Northeastern Europe 2005, Northern Subregion

找出树的所有重心，其实把1655的一个重心改一下就可以。
思路是找到一个小于的就清空答案数组，等于的就pushback。

流输入输出会超时，数组小了会RE。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 50005
using namespace std;

int cnt;
int t,n,root;
bool vis[MAXN];
int head[MAXN],sa[MAXN];
int son[MAXN];
int p=1;
int ans=1<<30,ansid;

struct point{
    int to,next;
}e[2*MAXN];



inline void add(int x,int y){
    e[++cnt].to = y;
    e[cnt].next = head[x];
    head[x]=cnt;
}

void dfs(int id){
    vis[id]=1;
    int tmp=0;
    son[id]=1;
    for(register int i=head[id];i!=-1;i=e[i].next ){
        int v=e[i].to ;
        if(vis[v]) continue;
//      vis[v]=1;
        dfs(v);
        son[id]+=son[v];
        tmp=max(son[v],tmp);
    }
    //tmp is the max number of sub trees
    tmp=max(tmp,n-son[id]);
//  tmp=max(tmp,n-son[id]+1);
    if(tmp<=ans){//写成小于号了! 
//      cout<<id<<" !
";
        if(tmp==ans){
            sa[++p] = id;
//          sa[p].siz = tmp;
        }
        else{
            p=1;
            sa[p]= id;
//          sa[p].siz = tmp;
            ans=tmp;    
        }

    }
}

int main(){
    int x,y;
        cnt=0;//!!
        memset(head,-1,sizeof(head));
//        memset(son,0,sizeof(son));//!!!
//        memset(vis,0,sizeof(vis));//!!!
//        ans=1<<30;
//        ansid=0;
        cin>>n;
        for(register int i=1;i<n;i++){
            scanf("%d%d",&x,&y);
            add(x,y);
            add(y,x);
        }

//      while(book[i]) i++;
//      root=i;
//      cout<<root<<endl;
//      vis[root]=1;
        dfs(1);
        sort(sa+1,sa+1+p);
        for(register int i=1;i<=p;i++){
            printf("%d ",sa[i]);
        }

    return 0;
}