B. 0-1 MST

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Ujan has a lot of useless stuff in his drawers, a considerable part of which are his math notebooks: it is time to sort them out. This time he found an old dusty graph theory notebook with a description of a graph.

It is an undirected weighted graph on n vertices. It is a complete graph: each pair of vertices is connected by an edge. The weight of each edge is either 0 or 1; exactly m edges have weight 1, and all others have weight 0.

Since Ujan doesn’t really want to organize his notes, he decided to find the weight of the minimum spanning tree of the graph. (The weight of a spanning tree is the sum of all its edges.) Can you find the answer for Ujan so he stops procrastinating?

Input
The first line of the input contains two integers n and m (1≤n≤105, 0≤m≤min(n(n−1)2,105)), the number of vertices and the number of edges of weight 1 in the graph.

The i-th of the next m lines contains two integers ai and bi (1≤ai,bi≤n, ai≠bi), the endpoints of the i-th edge of weight 1.

It is guaranteed that no edge appears twice in the input.

Output
Output a single integer, the weight of the minimum spanning tree of the graph.

Examples
inputCopy
6 11
1 3
1 4
1 5
1 6
2 3
2 4
2 5
2 6
3 4
3 5
3 6
outputCopy
2
inputCopy
3 0
outputCopy
0
Note
The graph from the first sample is shown below. Dashed edges have weight 0, other edges have weight 1. One of the minimum spanning trees is highlighted in orange and has total weight 2.

很明显就是求补图的连通分量的个数。

这是一次cf的裸题。

我们肯定不能存补图，因为太大了，我们可以从能否到达来考虑。

用set维护所有点，先set存所有点。

每次循环，如果当前的这个点没有被染色，那么就从这个点开始bfs，找到所有当前点不能到的点，这些点是肯定能用长度为0的边连接的，可以划分为一块。然后删除这些块中的点。

AC代码：

#pragma GCC optimize(2)
#include<bits/stdc++.h>
//#define int long long
using namespace std;
const int N=1e5+10;
int n,m,cnt,vis[N];
set<int> st,g[N];
void bfs(int s,int col){
	queue<int> q;	q.push(s);	st.erase(s);
	while(q.size()){
		int u=q.front();	q.pop(); vector<int> v; vis[u]=col;
		for(set<int>::iterator it=st.begin();it!=st.end();it++){
			if(g[u].find(*it)==g[u].end())	v.push_back(*it),q.push(*it);
		}
		for(int i=0;i<v.size();i++)	st.erase(v[i]),vis[v[i]]=col;
	}
}
signed main(){
	cin>>n>>m;
	for(int i=1,a,b;i<=m;i++)	scanf("%d %d",&a,&b),g[a].insert(b),g[b].insert(a);
	for(int i=1;i<=n;i++)	st.insert(i);
	for(int i=1;i<=n;i++)	if(!vis[i])	bfs(i,++cnt);
	cout<<cnt-1<<endl;
	return 0;
}