令dp[u]表示，以u为节点，向下走的路径数目。一次dfs能够做完。再令f[u]表示，以u为节点，向上走需要的代价。
对于u的子节点v，计算v向下走的代价，计算v向上走的代价。
向上走的代价=(u去掉v向下走的代价+f[u])
所以f[v]=(u去掉v向下走的代价）+f[u]
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#pragma GCC optimize(2)
#define up(i,a,b) for(int i=a;i<b;i++)
#define dw(i,a,b) for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
char ch = getchar(); ll x = 0, f = 1;
while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 2e5 + 10;
struct eg {
int to, next;
}edge[N<<1];
int head[N << 1];
int val[N << 1];
int sz[N];
int fa[N];
vector<int>ans[N];
int cnt;
int dp[N];
void addedge(int u, int v)
{
edge[cnt].to = v;
edge[cnt].next = head[u];
val[cnt] = 1;
head[u] = cnt++;
edge[cnt].to = u;
edge[cnt].next = head[v];
head[v] = cnt++;
}
void dfs(int u, int f)
{
fa[u] = f;
sz[u] = 1;
for (int i = head[u]; ~i; i = edge[i].next)
{
int v = edge[i].to;
if (v == f)continue;
dfs(v, u);
if (val[i])
dp[u] += dp[v] + 1;
else dp[u] += dp[v];
sz[u] += sz[v];
}
}
int f[N];
void dfs2(int u)
{
for (int i = head[u]; ~i; i = edge[i].next)
{
int v = edge[i].to;
if (v == fa[u])continue;
int cost;
if (val[i])
cost = sz[v] - 1 - dp[v] + sz[u] - sz[v] - 1 - (dp[u] - dp[v] - 1) + 1 + f[u];
else cost = sz[v] - 1 - dp[v] + sz[u] - sz[v] - 1 - (dp[u] - dp[v]) + f[u];
if (val[i])
f[v] = 1 + sz[u] - sz[v] - 1 - (dp[u] - dp[v] - 1) + f[u];
else f[v] = sz[u] - sz[v] - 1 - (dp[u] - dp[v]) + f[u];
ans[cost].push_back(v);
dfs2(v);
}
}
int n;
int main()
{
n = read();
memset(head, -1, sizeof(head));
int u, v;
up(i, 0, n - 1)
{
u = read(), v = read();
addedge(u, v);
}
dfs(1, 0);
int cost_1 = n - 1 - dp[1];
ans[cost_1].push_back(1);
dfs2(1);
up(i,0, n)
{
if (!ans[i].empty())
{
sort(ans[i].begin(), ans[i].end());
cout << i << endl;
for (auto k : ans[i])
cout << k << " ";
break;
}
}
return 0;
}