[CF920E] Connected Components?

题目链接：Connected Components?

Description

一句话题意：求一张图的补图的连通块数。
给定一张 (n) 个点，(frac{n imes (n-1)}{2}-m) 条边的无向图。
读入 (m) 对点，表示不存在 (u) 到 (v) 这条边。
问这张图中有多少个连通块，并且将连通块的个数按不降序输出。
数据范围 (1le nle 200000, 0le mle min(frac{n imes (n-1)}{2}, 200000))

Solution

由抽屉原理知，必定存在一个点，与它相关的删去的边不超过(frac{m}{n})条。
我们找到这个点，然后将所有与它存在连边的点相连。显然，此时只剩下(frac{m}{n})个点还没有被匹配过。
对于剩下这些点，我们暴力枚举它们所连向的边即可。
复杂度 (O(frac{m}{n} imes n) = O(n))

Code

// Author: wlzhouzhuan
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define ull unsigned long long
#define rint register int
#define rep(i, l, r) for (rint i = l; i <= r; i++)
#define per(i, l, r) for (rint i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)

inline int read() {
  int x = 0, neg = 1; char op = getchar();
  while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
  while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
  return neg * x;
}
inline void print(int x) {
  if (x < 0) { putchar('-'); x = -x; }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}

const int N = 200005;
vector <int> adj[N];
int deg[N], n, m;

int f[N], sz[N], vis[N];
int find(int x) {
  return f[x] == x ? x : f[x] = find(f[x]);
}
void Union(int x, int y) {
  x = find(x), y = find(y);
  if (x != y) {
    sz[y] += sz[x];
    f[x] = y;
  }
}
void add(int x) {
  for (auto v: adj[x]) {
    vis[v] = 1;
  }
}
void del(int x) {
  for (auto v: adj[x]) {
    vis[v] = 0;
  }
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; i++) f[i] = i, sz[i] = 1;
  for (int i = 1; i <= m; i++) {
    int u = read(), v = read();
    adj[u].pb(v);
    adj[v].pb(u);
    deg[u]++, deg[v]++;
  }
  int ch = min_element(deg + 1, deg + n - 1) - deg;
  add(ch);
  vector <int> unc;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      Union(ch, i);
    } else {
      unc.push_back(i);
    }
  }
  del(ch);
  for (auto i: unc) {
    add(i);
    for (int j = 1; j <= n; j++) {
      if (!vis[j]) {
        Union(i, j);
      }
    }
    del(i);
  }
  
  vector <int> block;
  for (int i = 1; i <= n; i++) if (find(i) == i) {
    block.push_back(sz[find(i)]);
  }
  sort(block.begin(), block.end());
  printf("%d
", block.size());
  for (auto v: block) {
    printf("%d ", v); 
  } 
  puts("");
  return 0;
}