奶牛跨栏(Cow Hurdles 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, m, t;
    if (!(cin >> n >> m >> t)) return 0;
    const int INF = 1000000007;
    vector<vector<int>> d(n + 1, vector<int>(n + 1, INF));
    for (int i = 1; i <= n; ++i) d[i][i] = 0;
    for (int i = 0; i < m; ++i) {
        int u, v, h;
        cin >> u >> v >> h;
        d[u][v] = min(d[u][v], h);
    }
    for (int k = 1; k <= n; ++k) {
        for (int i = 1; i <= n; ++i) {
            if (d[i][k] >= INF) continue;
            for (int j = 1; j <= n; ++j) {
                d[i][j] = min(d[i][j], max(d[i][k], d[k][j]));
            }
        }
    }
    for (int i = 0; i < t; ++i) {
        int a, b;
        cin >> a >> b;
        cout << (d[a][b] >= INF ? -1 : d[a][b]) << "\n";
    }
    return 0;
}