解法一：并查集

建立关系网络，用 dfs 实现并查集，统计人数和 weight 是否满足要求，最后按照名字的字典序排列输出。

#include <bits/stdc++.h>
using namespace std;
class Person {
public:
    string name;
    int weight;
    bool vis;
    set<Person *> next;
    Person() {}
    Person(string name, int weight) : name(name), weight(weight), vis(false) {}
};
map<string, Person> gangMap;
int cnt;
int total;
Person *gang;
vector<pair<Person *, int>> ans;
void dfs(Person &p) {
    ++cnt;
    total += p.weight;
    if (p.weight > gang->weight) {
        gang = &p;
    }
    p.vis = true;
    for (auto &it:p.next) {
        if (!it->vis) {
            dfs(*it);
        }
    }
}
bool cmp(pair<Person *, int> &o1, pair<Person *, int> &o2) {
    return o1.first->name < o2.first->name;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int N, K;
    cin >> N >> K;
    string name1, name2;
    int time;
    map<string, Person>::iterator res1, res2;
    for (int i = 0; i < N; ++i) {
        cin >> name1 >> name2 >> time;
        res1 = gangMap.find(name1);
        if (res1 == gangMap.end()) {
            gangMap.emplace(name1, Person(name1, time));
        } else {
            res1->second.weight += time;
        }
        res2 = gangMap.find(name2);
        if (res2 == gangMap.end()) {
            gangMap.emplace(name2, Person(name2, time));
        } else {
            res2->second.weight += time;
        }
        res1 = gangMap.find(name1);
        res2 = gangMap.find(name2);
        res1->second.next.emplace(&res2->second);
        res2->second.next.emplace(&res1->second);
    }
    for (auto &it:gangMap) {
        if (!it.second.vis) {
            cnt = 0;
            total = 0;
            gang = &it.second;
            dfs(it.second);
            if (cnt > 2 && total / 2 > K) {
                ans.emplace_back(make_pair(gang, cnt));
            }
        }
    }
    sort(ans.begin(), ans.end(), cmp);
    cout << ans.size() << '\n';
    for (auto &it:ans) {
        cout << it.first->name << " " << it.second << '\n';
    }
}