采购特价商品(Special Offer Shopping) 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 <cmath>
#include <iomanip>
#include <algorithm>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<double> x(n + 1), y(n + 1);
    for (int i = 1; i <= n; ++i) cin >> x[i] >> y[i];
    int m;
    cin >> m;
    const double INF = 1e18;
    vector<vector<double>> d(n + 1, vector<double>(n + 1, INF));
    for (int i = 1; i <= n; ++i) d[i][i] = 0.0;
    for (int i = 0; i < m; ++i) {
        int u, v;
        cin >> u >> v;
        double w = hypot(x[u] - x[v], y[u] - y[v]);
        d[u][v] = min(d[u][v], w);
        d[v][u] = min(d[v][u], w);
    }
    for (int k = 1; k <= n; ++k) {
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= n; ++j) {
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
            }
        }
    }
    int s, t;
    cin >> s >> t;
    cout << fixed << setprecision(2) << d[s][t] << "\n";
    return 0;
}