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poj3107（樹形dp）

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poj3107（樹形dp）

Godfather Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 5586 Accepted: 1950

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

Sample Output

2 3

Source

Northeastern Europe 2005, Northern Subregion
題意：給定一顆n個節點的樹，刪除其中一個節點，樹被分為很多塊，分出來塊最小的就是Godfather；求所有的這樣的Godfather，按升序輸出。分析：刪除一個節點，分成的塊就是該點的父結點所在的塊和其所有子節點的塊，每次找最大的塊，然後在找出的所有塊中找出最小的塊。

#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#include 
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define MAXN 100010

struct node
{
    int v;
    node *next;
}tree[MAXN], *head[MAXN];
int pre,n,dp[MAXN];//dp用來裝每個結點的子節點數
int num[MAXN],ans,tot;//num用來裝最後的答案

void init()//初始化
{
    pre = 1; ans = INF;
    CL(dp, 0);
    CL(head, NULL);
}

void add(int a, int b)//加邊，即建樹
{
    tree[pre].v = a;
    tree[pre].next = head[b]; head[b] = &tree[pre++];
    tree[pre].v = b;
    tree[pre].next = head[a]; head[a] = &tree[pre++];
}

void dfs(int son, int father)//記錄每個結點的子節點數
{
    node *p = head[son];
    dp[son] = 1;
    while(p != NULL)
    {
        if(p->v != father)
        {
            dfs(p->v, son);
            dp[son] += dp[p->v];
        }
        p = p->next;
    }
}

void dp_tree(int son, int father)
{
    int maxx = 0;
    node *p = head[son];
    while(p != NULL)
    {
        if(p->v != father)
        {
            dp_tree(p->v, son);
            maxx = max(dp[p->v], maxx);
        }
        p = p->next;
    }
    //更新答案
    maxx = max(n-dp[son], maxx);
    if(maxx < ans)
    {
        tot = 1;
        ans = maxx;
        num[tot] = son;
    }
    else if(maxx == ans)
    {
        tot++;
        num[tot] = son;
    }
}

int main()
{
    int u,v;
    while(scanf(%d,&n)==1)
    {
        init();
        for(int i=1; i