CF219D

/*
*题目大意：
*        在一棵有向树上，使任意一个点为cap,满足从这个cap能够到达所有点，
*        条件是是你可以更改边的方向。求所有改边最小的cap.
*解题思路：
*        一开始直接暴力肯定是TLE,然后想到了可以从入度为0的点开始考虑，
*        复杂度开始下降，但是明显随即数据很难搞定，接着想到了是不是
*        重复计算次数太多了。如何减少重复计算？想到这里就没有再细想，
*        没想到恰恰这里就是关键，先计算好随意一个cap点的改边数量。之后
*        其它的点均可以由这个点来推算出。
*/

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <utility>
#include <memory.h>
#include <stack>
#include <queue>
#include <deque>
#include <time.h>
#include <iomanip>

#define pi 3.1415926535897932
#define abs(x) ((x)>0?(x):-(x))
#define sqr(x) ((x)*(x))
#define sci(x) scanf("%d",&x)
#define scd(x) scanf("%lf",&x)
#define pri(x) printf("%d",x)
#define prd(x) printf("%lf",x)
#define inf 0x3f3f3f3f
#define fi first
#define se second
#define tr(container, iterator) for(typeof(container.begin()) iterator=container.begin(); it != container.end(); it++)
#define all(x) x.begin(), x.end()
#define y0 stupid_cmath
#define y1 very_stupid_cmath
#define ll long long
#define pb push_back
#define mp make_pair
#define rep(it,a,b) for(int it=a;it<=b;++it)
#define _rep(it,a,b) for(int it=a;it>=b;--it)

using namespace std;

const int MAXN = 200005;
const int MAXE = 200005;
typedef struct _node
{
    int v, w;
    int next;
}N;

vector<int> ans;
int Min;
N edge[2 * MAXE];
int cntEdge, head[MAXN];
int dp[MAXN];

void init(int n)
{
    Min = 0x7fffffff;
    cntEdge = 0;
    fill(dp, dp + n + 1, 0);
    fill(head, head + n + 1, -1);
}

void addEdge(int u, int v)
{
    edge[cntEdge].v = v;
    edge[cntEdge].w = 0;
    edge[cntEdge].next = head[u];
    head[u] = cntEdge++;

    edge[cntEdge].v = u;
    edge[cntEdge].w = 1;
    edge[cntEdge].next = head[v];
    head[v] = cntEdge++;
}

void dfs1(int n, int p)
{
    for(int f = head[n]; f != -1; f = edge[f].next)
    {
        if(edge[f].v != p)
        {
            if(edge[f].w)
                dp[1]++;
            dfs1(edge[f].v, n);
        }
    }
}

void dfs2(int n, int p)
{
    for(int f = head[n]; f != -1; f = edge[f].next)
    {
        if(edge[f].v != p)
        {
            if(edge[f].w == 0)
                dp[edge[f].v] = dp[n] + 1;
            else
                dp[edge[f].v] = dp[n] - 1;
            dfs2(edge[f].v, n);
        }
    }
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif

    int n;
    while(scanf("%d", &n) == 1)
    {
        init(n);
        int u, v;
        for(int i = 0; i < n - 1; i++)
        {
            scanf("%d %d", &u, &v);
            addEdge(u, v);
        }
        
        dfs1(1, 1);
        dfs2(1, 1);
        
        rep(i, 1, n)
            if(dp[i] < Min)
            {
                Min = dp[i];
                ans.clear();
                ans.push_back(i);
            }
            else if(dp[i] == Min)
            {
                ans.push_back(i);
            }

        printf("%d\n", Min);
        sort(ans.begin(), ans.end());
        rep(i, 0, ans.size() - 1)
        {
            if(!i)
                printf("%d", ans[i]);
            else
                printf(" %d", ans[i]);
        }
        puts("");
    }
    return 0;
}