[Cf1243D] D.0-1 MST (set + shared equally complexity analysis)

Portal

That Italy:
an existing \ (n, n \ leq 10 ^ 5 \) completely nodes of FIG.
Given \ (m, m \ leq 10 ^ 5 \) Article \ (1 \) side, the remaining sides are all \ (0 \) side.
Now requires that this figure \ (MST \) right is how much.

Ideas:

• Obviously there is a violent idea: use a queue has been maintained in (MST \) \ node in, then we direct enumeration of all the \ (0 \) side can be judged and then into the team. Finally determined the \ (X \) a communication block, then the answer is \ (. 1-X \) .
• But here \ (0 \) the number of edges we can not be too many to enumerate. We note that the practice of violence are "add" side, we can consider the reverse, namely the "delete" side: We enumerate each \ (1 \) side and retain these nodes, all the remaining nodes directly into the team.
• Here we use a \ (set \) to maintain the remaining points are not added, so you can easily remove them from the junction.
• A point will be out \ (set \) and queue up again, a side only two points, a point that is most remain in the \ (set \) inside twice. So the overall complexity is \ (O (nlogn) \) .

I code for convenience of the adjacency matrix directly, to keep a map, so that the complexity of a more \ (log \) .
code show as below:

/*
 * Author:  heyuhhh
 * Created Time:  2020/3/5 16:55:05
 */
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
  #define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
  void err() { std::cout << '\n'; }
  template<typename T, typename...Args>
  void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
  template <template<typename...> class T, typename t, typename... A> 
  void err(const T <t> &arg, const A&... args) {
  for (auto &v : arg) std::cout << v << ' '; err(args...); }
#else
  #define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 1e5 + 5;

int n, m;
map <int, bool> mp[N];
set <int> s;
queue <int> q;

void run() {
    cin >> n >> m;
    for(int i = 1; i <= m; i++) {
        int u, v; cin >> u >> v;
        mp[u][v] = mp[v][u] = true;
    }
    for(int i = 1; i <= n; i++) s.insert(i);
    int t = 0;
    while(!s.empty()) {
        auto tmp = s.begin();
        q.push(*tmp);
        s.erase(tmp);
        ++t;
        while(!q.empty()) {
            int u = q.front(); q.pop();
            for(auto it = s.begin(), nxt = it; it != s.end(); it = nxt) {
                int v = *it;
                ++nxt;
                if(mp[u].find(v) == mp[u].end()) {
                    q.push(v);
                    s.erase(it);   
                }
            }           
        }
    }
    cout << t - 1 << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cout << fixed << setprecision(20);
    run();
    return 0;
}