Codeforces Round #599 (Div. 2)
D. 0-1 MST
Description
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.
input1
6 11
1 3
1 4
1 5
1 6
2 3
2 4
2 5
2 6
3 4
3 5
3 6
output1
2
Meaning of the questions:
To give you a complete graph of n, wherein m is 1 weights edges, the other side of the weight is 0. Size calculated minimum weight spanning tree weights.
answer
We set of n points into the vessel and to build a set of FIG container, then bfs violence, obtains the number of connected blocks, the number of blocks communicated answer is -1.
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N=1e5+10;
set<int>G[N],s;
int vis[N];
void bfs(int x)
{
queue<int>q;
q.push(x);
s.erase(x);
while(q.size()>0)
{
int y=q.front();
q.pop();
if(vis[y])
continue;
vis[y]=1;
for(auto it=s.begin();it!=s.end();)
{
int v=*it;
++it;
if(G[y].find(v)==G[y].end())
{
q.push(v);//cout<<"-";
s.erase(v);
}
}
}
}
int main()
{
int n,m;
cin>>n>>m;
for(int i=1;i<=n;i++)
{
s.insert(i);
}
for(int i=1;i<=m;i++)
{
int x,y;
cin>>x>>y;
G[x].insert(y);
G[y].insert(x);
}
int ans=0;
for(int i=1;i<=n;i++)
{
if(!vis[i])
{
bfs(i);
ans++;
}
}
cout<<ans-1<<"\n";
return 0;
}