树上双字符连通块划分(We Need Both a and b) 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;

const long long MOD = 1000000007LL;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<char> c(n + 1);
    for (int i = 1; i <= n; ++i) cin >> c[i];
    vector<vector<int>> adj(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;
                order.push_back(v);
            }
        }
    }
    // dp0[u]: only 'a', dp1[u]: only 'b', dp2[u]: both 'a' and 'b'
    vector<long long> dp0(n + 1, 0), dp1(n + 1, 0), dp2(n + 1, 0);
    vector<long long> prodSingle(n + 1, 1), prodAll(n + 1, 1);
    for (int i = n - 1; i >= 0; --i) {
        int u = order[i];
        if (c[u] == 'a') {
            dp0[u] = prodSingle[u];
            dp2[u] = (prodAll[u] - prodSingle[u] + MOD) % MOD;
        } else {
            dp1[u] = prodSingle[u];
            dp2[u] = (prodAll[u] - prodSingle[u] + MOD) % MOD;
        }
        int p = par[u];
        if (p) {
            long long same = (c[p] == 'a' ? dp0[u] : dp1[u]);
            prodSingle[p] = prodSingle[p] * ((same + dp2[u]) % MOD) % MOD;
            prodAll[p] = prodAll[p] * ((dp0[u] + dp1[u] + 2LL * dp2[u]) % MOD) % MOD;
        }
    }
    cout << dp2[1] << "\n";
    return 0;
}