贪婪的 Copy(Greedy Copy) 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<vector<long long>> d(n + 1, vector<long long>(n + 1));
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            cin >> d[i][j];
        }
    }
    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 p;
    cin >> p;
    vector<int> t(p);
    for (int i = 0; i < p; ++i) cin >> t[i];
    if (p == 0) {
        cout << d[1][n] << "\n";
        return 0;
    }
    const long long INF = (1LL << 58);
    int lim = 1 << p;
    vector<vector<long long>> dp(lim, vector<long long>(p, INF));
    for (int i = 0; i < p; ++i) {
        dp[1 << i][i] = d[1][t[i]];
    }
    for (int mask = 1; mask < lim; ++mask) {
        for (int i = 0; i < p; ++i) {
            if (dp[mask][i] >= INF) continue;
            for (int j = 0; j < p; ++j) {
                if (!(mask & (1 << j))) {
                    int nmask = mask | (1 << j);
                    dp[nmask][j] = min(dp[nmask][j], dp[mask][i] + d[t[i]][t[j]]);
                }
            }
        }
    }
    long long ans = INF;
    for (int i = 0; i < p; ++i) {
        ans = min(ans, dp[lim - 1][i] + d[t[i]][n]);
    }
    cout << ans << "\n";
    return 0;
}