重建道路(Rebuilding Roads) 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>
#include <algorithm>
using namespace std;

const int INF = 1000000;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, P;
    if (!(cin >> n >> P)) return 0;
    vector<vector<int>> children(n + 1);
    for (int i = 1; i < n; ++i) {
        int u, v;
        cin >> u >> v;
        children[u].push_back(v);
    }
    vector<int> order;
    order.push_back(1);
    for (int idx = 0; idx < (int)order.size(); ++idx) {
        int u = order[idx];
        for (int v : children[u]) order.push_back(v);
    }
    vector<int> sz(n + 1, 1);
    vector<vector<int>> dp(n + 1, vector<int>(n + 1, INF));
    for (int i = n - 1; i >= 0; --i) {
        int u = order[i];
        dp[u][1] = (int)children[u].size();
        for (int v : children[u]) {
            for (int j = sz[u] + sz[v]; j >= 1; --j) {
                for (int k = max(1, j - sz[u]); k <= min(j - 1, sz[v]); ++k) {
                    dp[u][j] = min(dp[u][j], dp[u][j - k] + dp[v][k] - 1);
                }
            }
            sz[u] += sz[v];
        }
    }
    int ans = dp[1][P];
    for (int i = 2; i <= n; ++i) {
        if (sz[i] >= P) {
            ans = min(ans, dp[i][P] + 1);
        }
    }
    cout << ans << "\n";
    return 0;
}