回家(Bessie Come Home) 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 charId(char c) {
    if (c >= 'A' && c <= 'Z') return c - 'A';
    return 26 + (c - 'a');
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int p;
    if (!(cin >> p)) return 0;
    const long long INF = (1LL << 58);
    vector<vector<long long>> d(52, vector<long long>(52, INF));
    for (int i = 0; i < 52; ++i) d[i][i] = 0;
    for (int i = 0; i < p; ++i) {
        char a, b;
        long long w;
        cin >> a >> b >> w;
        int u = charId(a), v = charId(b);
        d[u][v] = min(d[u][v], w);
        d[v][u] = min(d[v][u], w);
    }
    for (int k = 0; k < 52; ++k) {
        for (int i = 0; i < 52; ++i) {
            for (int j = 0; j < 52; ++j) {
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
            }
        }
    }
    int best_c = 0;
    long long best_d = INF;
    for (int i = 0; i < 25; ++i) {
        if (d[i][25] < best_d) {
            best_d = d[i][25];
            best_c = i;
        }
    }
    cout << char('A' + best_c) << " " << best_d << "\n";
    return 0;
}