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
題意：
給你一個n個點的完全圖，其中m條邊權值是1，其他邊的權值是0。求出權值最小的生成樹權值的大小。
題解
我們把n個點放到set容器中和建一個set容器的圖，然后通過bfs暴力，求出連通塊的個數，答案就是連通塊的個數-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;
}