子树染色计数(Subtree) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    long long M;
    if (!(cin >> n >> M)) return 0;
    vector<vector<int>> adj(n + 1), ch(n + 1);
    for (int i = 1; i < n; ++i) {
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    vector<int> par(n + 1, 0), order;
    vector<bool> vis(n + 1, false);
    vis[1] = true;
    order.push_back(1);
    for (int idx = 0; idx < (int)order.size(); ++idx) {
        int u = order[idx];
        for (int v : adj[u]) {
            if (!vis[v]) {
                vis[v] = true;
                par[v] = u;
                ch[u].push_back(v);
                order.push_back(v);
            }
        }
    }
    vector<long long> down(n + 1, 1), up(n + 1, 1);
    for (int i = n - 1; i >= 0; --i) {
        int u = order[i];
        for (int v : ch[u]) {
            down[u] = down[u] * ((down[v] + 1) % M) % M;
        }
    }
    for (int i = 0; i < n; ++i) {
        int u = order[i];
        int k = (int)ch[u].size();
        vector<long long> pref(k + 1, 1), suff(k + 1, 1);
        for (int j = 0; j < k; ++j) {
            pref[j + 1] = pref[j] * ((down[ch[u][j]] + 1) % M) % M;
        }
        for (int j = k - 1; j >= 0; --j) {
            suff[j] = suff[j + 1] * ((down[ch[u][j]] + 1) % M) % M;
        }
        for (int j = 0; j < k; ++j) {
            int v = ch[u][j];
            long long withoutV = pref[j] * suff[j + 1] % M * up[u] % M;
            up[v] = (withoutV + 1) % M;
        }
    }
    for (int i = 1; i <= n; ++i) {
        cout << (down[i] * up[i]) % M << "\n";
    }
    return 0;
}