POJ3107--Godfather（树的重心）

vector建图被卡了。。改为链式前向星500ms过的。。差了四倍多？。。。

表示不太会用链表建图啊。。自己试着写的，没看模板。。嗯。。果然错了。。落了一句话orz

树的重心就是找到一个树中一个点,其所有的子树中最大的子树节点数最少,那么这个点就是这棵树的重心,删去重心后，生成的多棵树尽可能平衡。

根据定义很容易求得。

/***********************************************
Problem: 3107		User: G_lory
Memory: 4440K		Time: 500MS
Language: C++		Result: Accepted
***********************************************/

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <complex>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <cassert>
using namespace std;
typedef long long ll;

const int N = 500005;

struct Edge {
    int to, next;
} edge[N];
int head[N];

int cnt;
int sz[N];
int maxn[N];

int n;

void add_edge(int u, int v) // 双向边
{
    edge[cnt].to = v;
    edge[cnt].next = head[u];
    head[u] = cnt;
    cnt++;

    edge[cnt].to = u;
    edge[cnt].next = head[v];
    head[v] = cnt;
    cnt++;
}

int ans;
void dfs(int u, int pre)
{
    sz[u] = 0;
    int temp = 0;
    for (int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].to;
        if (v == pre) continue;
        dfs(v, u);
        sz[u] += sz[v] + 1;
        temp = max(temp, sz[v] + 1);
    }
    temp = max(n - sz[u] - 1, temp);
    maxn[u] = temp;
    ans = min(temp, ans);
}

int main()
{
    while (~scanf("%d", &n))
    {
        int a, b;
        memset(head, -1, sizeof head);
        cnt = 1;
        for (int i = 1; i < n; ++i)
        {
            scanf("%d%d", &a, &b);
            add_edge(a, b);
        }
        ans = n;
        dfs(1, -1);
        int ok = 1;
        for (int i = 1; i <= n; ++i)
        {
            if (maxn[i] == ans)
            {
                if (ok) ok = 0;
                else printf(" ");
                printf("%d", i);
            }
        }
        printf("\n");
    }
    return 0;
}