逃离牛棚(Running Away From the Barn 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;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    long long t;
    if (!(cin >> n >> t)) return 0;
    const int LOG = 19;
    vector<vector<int>> up(LOG, vector<int>(n + 1, 0));
    vector<long long> dist(n + 1, 0);
    for (int i = 2; i <= n; ++i) {
        int p;
        long long w;
        cin >> p >> w;
        up[0][i] = p;
        dist[i] = dist[p] + w;
        for (int k = 1; k < LOG; ++k) {
            up[k][i] = up[k - 1][up[k - 1][i]];
        }
    }
    vector<int> diff(n + 1, 0);
    for (int i = 1; i <= n; ++i) {
        int cur = i;
        for (int k = LOG - 1; k >= 0; --k) {
            int anc = up[k][cur];
            if (anc != 0 && dist[i] - dist[anc] <= t) {
                cur = anc;
            }
        }
        diff[i]++;
        if (up[0][cur]) diff[up[0][cur]]--;
    }
    for (int i = n; i >= 2; --i) {
        diff[up[0][i]] += diff[i];
    }
    for (int i = 1; i <= n; ++i) cout << diff[i] << "\n";
    return 0;
}