谷仓涂色(Barn Painting G) 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, k;
    if (!(cin >> n >> k)) return 0;
    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> fixedCol(n + 1, 0);
    for (int i = 0; i < k; ++i) {
        int b, c;
        cin >> b >> c;
        fixedCol[b] = c;
    }
    vector<vector<long long>> dp(n + 1, vector<long long>(4, 1));
    for (int i = 1; i <= n; ++i) {
        if (fixedCol[i] != 0) {
            for (int c = 1; c <= 3; ++c) {
                if (c != fixedCol[i]) dp[i][c] = 0;
            }
        }
    }
    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);
            }
        }
    }
    for (int i = n - 1; i >= 0; --i) {
        int u = order[i];
        int p = par[u];
        if (p) {
            dp[p][1] = dp[p][1] * ((dp[u][2] + dp[u][3]) % MOD) % MOD;
            dp[p][2] = dp[p][2] * ((dp[u][1] + dp[u][3]) % MOD) % MOD;
            dp[p][3] = dp[p][3] * ((dp[u][1] + dp[u][2]) % MOD) % MOD;
        }
    }
    cout << (dp[1][1] + dp[1][2] + dp[1][3]) % MOD << "\n";
    return 0;
}