Godfather POJ - 3107

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

网传都是dp。。。- -！

我怎么感觉这不是个dp呢

首先处理一下这个图，从任意一点，开始向着一个方向深入，求出每一个子树的子节点数

假设某一个节点的子树有n个，那么这个点就把图分成了n+1个子树，其中的1是父亲节点的树。

注意 STL超时，我试过了，前向星能过

STL果断超时了

#include
#include
#include
using namespace std;
const int M = 50000+10;
vectoredge[M];
int tol[M],n;
int ans[M],sans[M];
int minn;
int culson(int son,int f)
{
    tol[son]=1;
    int len=edge[son].size();
    for(int i=0;i

前向星：

#include
#include
#include
using namespace std;
const int M = 50000+10;
int tol[M],n,anss;
int ans[M],sans[M];
int pre[M];
int minn;
struct aa
{
    int v,next;
} edge[5*M];
void add(int u,int v)
{
    edge[anss].v=v;
    edge[anss].next=pre[u];
    pre[u]=anss++;
}
int culson(int son,int f)
{
    tol[son]=1;
    for(int i=pre[son]; i!=-1; i=edge[i].next)
    {
        int v=edge[i].v;
        if(v!=f)
        {
            tol[son]+=culson(v,son);
        }
    }
    return tol[son];
}
void update(int son,int f)
{
    ans[son]=0;
    for(int i=pre[son]; i!=-1; i=edge[i].next)
    {
        int v=edge[i].v;
        if(v!=f)
        {
            ans[son]=max(ans[son],tol[v]);
            update(v,son);
        }
    }
    ans[son]=max(ans[son],n-tol[son]);
    minn=min(ans[son],minn);
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        minn=0x3f3f3f3f;
        anss=0;
        int u,v;
        for(int i=1; i<=n; i++)
        {
            tol[i]=0;
            pre[i]=-1;
        }

        for(int i=0; i