遇见两个树形dp的题目，写一下

牛客小白月赛45-E

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const int maxn = 1e5 + 10;
ll arr[maxn];
ll res = -0x3f3f3f3f;
int mark[maxn];
struct node {
    int y;
    ll w;
    node(int a, ll b): y(a), w(b) {}
};
vector<node> v[maxn];
ll dfs(int rt, ll val) {
    ll ans = 0;

    for (auto to:v[rt]) {
        node pos = to;
        if (mark[pos.y])
            continue;
        mark[pos.y] = 1;
        ll z = dfs(pos.y, pos.w);
        if (z > 0)
            ans += z;
    }
    res = max(res, ans + arr[rt]);
    return ans + val + arr[rt];
}

int main() {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> arr[i];
        res = max(res, arr[i]);
    }
    int x, y;
    ll w;
    for (int i = 1; i < n; i++) {
        cin >> x >> y >> w;
        v[x].push_back(node(y, w));
        v[y].push_back(node(x, w));
    }
    mark[1] = 1;
    dfs(1, 0);
    cout << res << endl;
    return 0;
}

牛客练习赛97-D

树形dp，$dp[i][0/1]$表示染特殊颜色或普通颜色的方案数。转移方程:
$$
\left\{\begin{aligned}dp[i][0]&=\prod(dp[son][0]\cdot x+dp[son][1]\cdot y)\\dp[i][1]&=\prod(dp[son][0]\cdot x+dp[son][1]\cdot(y-1))\end{aligned}\right.
$$

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
constexpr ll mod = 998244353;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    ll x, y;
    cin >> n >> x >> y;
    vector<vector<int>> G(n + 1);  // G最多放n+1个向量
    for (int i = 1, u, v; i < n; i++) {
        cin >> u >> v;
        G[u].push_back(v);
        G[v].push_back(u);
    }
    function<pair<ll, ll>(int, int)> DFS = [&](int u, int p) {
        pair<ll, ll> res{1, 1};
        for (int v : G[u]) {
            if (v != p) {
                auto pr = DFS(v, u);
                res.first = (pr.first * x + pr.second * y) % mod * res.first % mod;
                res.second = (pr.first * x + pr.second * (y - 1)) % mod * res.second % mod;
            }
        }
        return res;
    };
    auto pr = DFS(1, 0);
    cout << (pr.first * x + pr.second * y) % mod << endl;
    // system("pause");
    return 0;
}