奶牛拜访(Visits S) 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;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<int> a(n + 1), indeg(n + 1, 0);
    vector<long long> v(n + 1);
    long long total = 0;
    for (int i = 1; i <= n; ++i) {
        cin >> a[i] >> v[i];
        indeg[a[i]]++;
        total += v[i];
    }
    vector<int> q;
    q.reserve(n);
    vector<int> rem(n + 1, 0);
    for (int i = 1; i <= n; ++i) {
        if (indeg[i] == 0) {
            q.push_back(i);
            rem[i] = 1;
        }
    }
    for (size_t head = 0; head < q.size(); ++head) {
        int u = q[head];
        int nxt = a[u];
        if (--indeg[nxt] == 0) {
            q.push_back(nxt);
            rem[nxt] = 1;
        }
    }
    for (int i = 1; i <= n; ++i) {
        if (!rem[i]) {
            long long mn = v[i];
            int cur = i;
            while (!rem[cur]) {
                rem[cur] = 1;
                mn = min(mn, v[cur]);
                cur = a[cur];
            }
            total -= mn;
        }
    }
    cout << total << "\n";
    return 0;
}