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?nn?vertices. It is a complete graph: each pair of vertices is connected by an edge. The weight of each edge is either?00?or?11; exactly?mm?edges have weight?11, and all others have weight?00.
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?nn?and?mm?(1≤n≤1051≤n≤105,?0≤m≤min(n(n?1)2,105)0≤m≤min(n(n?1)2,105)), the number of vertices and the number of edges of weight?11?in the graph.
The?ii-th of the next?mm?lines contains two integers?aiai?and?bibi?(1≤ai,bi≤n1≤ai,bi≤n,?ai≠biai≠bi), the endpoints of the?ii-th edge of weight?11.
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
Input
6 11
1 3
1 4
1 5
1 6
2 3
2 4
2 5
2 6
3 4
3 5
3 6
Output
2
Input
3 0
Output
0
Note
The graph from the first sample is shown below. Dashed edges have weight?00, other edges have weight?11. One of the minimum spanning trees is highlighted in orange and has total weight?22.
In the second sample, all edges have weight?00?so any spanning tree has total weight?00.
【思路】
要求最小生成树，如果考虑有0 边相连的点作为一个集合的话，那么就是求集合的个数。
求补图的连通块数减一，n个连通块需要n-1条边连接。
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。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
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;
}
}
}
int main()
{
ios::sync_with_stdio(false);
cin>>n>>m;
for(int i=1,a,b;i<=m;i++)
{
cin>>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;
}
代码是copydalao?的
cs