POJ 3107 Godfather(樹形dp)

畫船聽雨發表於2014-05-04
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題意是找大哥，是大哥的條件是：刪除這個節點之後最大聯通的點最少。

第一遍dfs找到每個節點有多少子節點，第二遍dfs求出刪掉每個環節點之後會剩餘多少個聯通的節點，並求出最大聯通節點的個數。

Godfather
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 4544   Accepted: 1539

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
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
//#define LL __int64
#define LL long long
#define INF 0x3f3f3f3f
#define PI 3.1415926535898

const int maxn = 50010;
using namespace std;

int head[maxn];
int t;
int Max;
int sum[maxn];
int p[maxn];
int n;
struct node
{
    int to;
    int next;
} f[2*maxn];

void init()
{
    t = 0;
    Max = INF;
    memset(head, -1, sizeof(head));
}

void add(int x, int y)
{
    f[t].to = y;
    f[t].next = head[x];
    head[x] = t++;

    f[t].to = x;
    f[t].next = head[y];
    head[y] = t++;
}

void dfs1(int x, int fa)
{
    sum[x] = 1;
    for(int i = head[x]; i != -1; i = f[i].next)
    {
        int v = f[i].to;
        if(v == fa)
            continue;
        dfs1(v, x);
        sum[x] += sum[v];
    }
}

void dfs2(int x, int fa)
{
    int &ans = p[x] = n-sum[x];
    for(int i = head[x]; i != -1; i = f[i].next)
    {
        int v = f[i].to;
        if(v == fa)
            continue;
        dfs2(v, x);
        ans = max(ans, sum[v]);
    }
    Max = min(Max, ans);
}

int main()
{
    while(~scanf("%d",&n))
    {
        int u, v;
        init();
        for(int i = 0; i < n-1; i++)
        {
            scanf("%d %d",&u, &v);
            add(u,v);
        }
        dfs1(1, -1);
        dfs2(1, -1);
        int flag = 0;
        for(int i = 1; i <= n; i++)
        {
            if(p[i] == Max)
            {
                if(!flag)
                {
                    flag = 1;
                    printf("%d",i);
                }
                else
                    printf(" %d",i);
            }
        }
        printf("\n");
    }
    return 0;
}


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