Submission

Score: 100

User: Nagornz

Problemset: อพยพปลอดภัย (Quartet)

Language: cpp

Time: 0.101 second

Submitted On: 2025-05-24 19:46:13

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

#define ordered_set <int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
#define ordered_multiset <int, null_type, less_equal <int>, rb_tree_tag, tree_order_statistics_node_update>

#define int long long
#define double long double
#define pii pair <int, int>
#define tiii tuple <int, int, int>
#define tiiii tuple <int, int, int, int>
#define emb emplace_back
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define iShowSpeed cin.tie(NULL)->sync_with_stdio(false)
#define matrix vector <vector <int>>
#define mat(n, m) vector <vector <int>> (n, vector <int> (m));

const int mod = 1e9 + 7;
const int inf = 1e18;
const matrix II = {{1, 0}, {0, 1}};
const int N = 1e5 + 5;
const int M = 1e5 + 5;

int n, m, sz[N + M];
vector <int> adj[N + M], sub[N + M];

void dfs(int u, int p){
    sz[u] = (u <= n);
    for (auto v : adj[u]) {
        if (v == p) continue;
        dfs(v, u);
        if (sz[v] > 0) sub[u].emb(sz[v]), sz[u] += sz[v];
    }
    sub[u].emb(n - sz[u]);
}

int32_t main(){
    iShowSpeed;
    cin >> n >> m;
    for (int i = 1; i < n + m; i++) {
        int u, v; cin >> u >> v;
        adj[u].emb(v);
        adj[v].emb(u);
    }
    dfs(1, 1);
    // for (int i = 1; i <= n + m; i++) {
    //     cout << i << ": ";
    //     for (auto e : sub[i]) cout << e << " ";
    //     cout << "\n";
    // }
    int ans = 0;
    for (int u = n + 1; u <= n + m; u++) {
        // cout << u << ": \n";
        int siz = sub[u].size();
        if (siz >= 4) {
            int dp[5]; memset(dp, 0, sizeof dp);
            dp[0] = 1;
            for (int i = 1; i <= siz; i++) {
                // cout << sub[u][i - 1] << "\n";
                for (int j = 4; j >= 1; j--) dp[j] = dp[j - 1] * sub[u][i - 1] + dp[j], dp[j] %= mod;
            }
            ans += dp[4]; ans %= mod;
        }
        // cout << "--------------------\n";
    }
    cout << ans;
}