每日一题 - 240322 - 快递分拣

• 快递分拣

---

• TAG - $芝士水题、算法-【STL-set、STL-map、STL-vector】$
• 时间复杂度 - $O(N)$

//
#include <bits/stdc++.h>
using namespace std;
// #define int long long 

const int N = 1e3 + 4;
set<string> st;
map<string, int> mp;
vector<string> ans_name;
vector<string> ans_id[N]; 

void solve() {
    int n;
    scanf("%d", &n);

    int idx = 0;
    while (n--) {
        string id, name;
        cin >> id >> name;

        if (st.find(name) == st.end()) {
            st.insert(name);
            ans_name.push_back(name);

            mp[name] = ++idx;
            ans_id[idx].push_back(id);
        } else {
            ans_id[mp[name]].push_back(id);
        }
    }

    for (int i = 0; i < ans_name.size(); i++) {
        cout << ans_name[i] << " " << ans_id[mp[ans_name[i]]].size() << endl;
        for (int j = 0; j < ans_id[mp[ans_name[i]]].size(); j++) {
            cout << ans_id[mp[ans_name[i]]][j] << endl;
        }
    }
}

signed main() {
    int t = 1;
    // scanf("%d", &t);
    while (t--) solve();
    return 0;
}

---

实现细节

• `

---

参考示意图

• `

---

参考链接

• `

---

作者 | 乐意奥AI