题干：

无向图中给定n个顶点，m条不存在的边(除了这m条边，其余都存在)，求图的连通分量，及每个连通分量的大小。

解题报告：

https://codeforces.com/blog/entry/4556

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<unordered_map>
#include<string>
#include<cmath>
#include<cstring>
#define F first
#define S second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 5e5 + 5;
unordered_map<int,bool> mp[MAX];
vector<int> vv;
vector<vector<int> > ans;
vector<int> bfs(int x) {vector<int> res;queue<int> q;q.push(x);while(q.size()) {int cur = q.front();q.pop();res.pb(cur);
//		int up = vv.size();这里不能这样写，因为size是动态的。for(int i = 0; i<(int)vv.size(); i++) {int v = vv[i];if(mp[cur].count(v) == 0) {swap(vv[i],vv.back());//用来实现set的erase操作，减少常数。 vv.pop_back();//用来实现set的erase操作 i--;q.push(v);}}}return res;
}
int main()
{int n,m;cin>>n>>m;for(int i = 1; i<=n; i++) vv.pb(i);for(int x,y,i = 1; i<=m; i++) {scanf("%d%d",&x,&y);mp[x][y]=1;mp[y][x]=1;}while(!vv.empty()) {int v = vv.back();vv.pop_back();vector<int> cnt = bfs(v);ans.push_back(cnt);}sort(ans.begin(),ans.end());printf("%d\n",ans.size());for(int i = 0; i<(int)ans.size(); i++) {printf("%d ",ans[i].size());for(int j = 0; j<(int)ans[i].size(); j++) {printf("%d ",ans[i][j]);}printf("\n");}return 0 ;
}