保安站岗(Security Guards) 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 long long INF = (1LL << 50);

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<long long> cost(n + 1, 0);
    vector<vector<int>> children(n + 1);
    vector<int> indeg(n + 1, 0);
    for (int i = 1; i <= n; ++i) {
        int u, m;
        long long k;
        cin >> u >> k >> m;
        cost[u] = k;
        for (int j = 0; j < m; ++j) {
            int v;
            cin >> v;
            children[u].push_back(v);
            indeg[v]++;
        }
    }
    int root = 1;
    for (int i = 1; i <= n; ++i) {
        if (indeg[i] == 0) {
            root = i;
            break;
        }
    }
    vector<int> order;
    order.push_back(root);
    for (int idx = 0; idx < (int)order.size(); ++idx) {
        int u = order[idx];
        for (int v : children[u]) order.push_back(v);
    }
    // dp0: placed at u, dp1: covered by child, dp2: covered by parent
    vector<long long> dp0(n + 1, 0), dp1(n + 1, 0), dp2(n + 1, 0);
    for (int i = n - 1; i >= 0; --i) {
        int u = order[i];
        dp0[u] = cost[u];
        dp2[u] = 0;
        long long extra = INF;
        for (int v : children[u]) {
            long long m01 = min(dp0[v], dp1[v]);
            dp0[u] += min(m01, dp2[v]);
            dp2[u] += m01;
            extra = min(extra, dp0[v] - m01);
        }
        dp1[u] = dp2[u] + extra;
    }
    cout << min(dp0[root], dp1[root]) << "\n";
    return 0;
}