題意：

給你一個圖，這個圖由兩個樹構成，問全域性最小割和方案數。

題解：

至少一個點度數不超過3，所以答案不超過3。

所以一定有一個樹只割了一條邊，列舉割的邊，看子樹和非子樹點在第二顆樹有多少個連邊就行了。用啟發式合併。

程式碼：

#include <bits/stdc++.h>
#ifdef LOCAL
#define debug(x) cout<<#x<<" = "<<(x)<<endl;
#else
#define debug(x) 1;
#endif

#define chmax(x,y) x=max(x,y)
#define chmin(x,y) x=min(x,y)
#define lson id<<1,l,mid
#define rson id<<1|1,mid+1,r
#define lowbit(x) x&-x
#define mp make_pair
#define pb push_back
#define fir first
#define sec second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const int MOD = 1e9 + 7;
const double PI = acos (-1.);
const double eps = 1e-10;
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int MAXN = 1e5 + 5;

struct Tree {
    vector<int> G[MAXN];
    int son[MAXN], siz[MAXN], n;

    void add (int x, int y) {
        G[x].pb (y);
    }

    void dfs (int now, int par) {
        siz[now] = 1;
        for (int i : G[now]) {
            if (i == par) continue;
            dfs(i, now);
            if (siz[son[now]] <= siz[i]) son[now] = i;
            siz[now] += siz[i];
        }
    }

} tr[2];

int vis[MAXN];
int out;

void cal (int now, Tree & b) {
    vis[now] ^= 1;
    for (int i : b.G[now]) {
        if (vis[i] != vis[now]) out++;
        else out--;
    }
}

void edt (int now, int par, Tree & a, Tree &b) {
    cal(now, b);
    for (int i : a.G[now]) if (i != par) edt(i, now, a, b);
}

pii ret;
void dfs (int now, int par, int kep, Tree &a, Tree & b) {
    for (int i : a.G[now]) {
        if (i == par || i == a.son[now]) continue;
        dfs(i, now, 0, a, b);
    }
    if (a.son[now]) dfs(a.son[now], now, 1, a, b);
    cal(now, b);
    for (int i : a.G[now]) if (i != par && i != a.son[now]) edt(i, now, a, b);
    if (par) {
        if (out < ret.first) ret.first = out, ret.second = 1;
        else if (out == ret.first) ret.second++;
    }
    if (!kep) edt(now, par, a, b);
}


void solve (Tree &a, Tree & b) {
    ret = {INF, 0};
    out = 0;
    a.dfs(1, 0);
    b.dfs(1, 0);
    dfs(1, 0, 0, a, b);
    if (ret.first == 1) {
        printf ("2 %d\n", ret.second);
    } else {
        out = 0;
        dfs(1, 0, 0, b, a);
        printf ("3 %d\n", ret.second);
    }
}



int main() {
#ifdef LOCAL
    freopen ("input.txt", "r", stdin);
#endif
    int n;
    scanf ("%d", &n);
    for (int i = 1; i < n; i++) {
        int x, y;
        scanf ("%d %d", &x, &y);
        tr[0].add(x, y);
        tr[0].add(y, x);
    }
    for (int i = 1; i < n; i++) {
        int x, y;
        scanf ("%d %d", &x, &y);
        tr[1].add(x, y);
        tr[1].add(y, x);
    }
    solve(tr[0], tr[1]);
    return 0;
}