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Published on 2017-09-02 11:31 in 暂未分类 with qscqesze

Codeforces Round #599 (Div. 1) B. 0-1 MST 图论

D. 0-1 MST

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

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 0, other edges have weight 1. One of the minimum spanning trees is highlighted in orange and has total weight 2.

In the second sample, all edges have weight 0 so any spanning tree has total weight 0.

题意

现在有n个点的完全图,其中m条边的长度是1,其他都是0;问你这个图的最小生成树的代价是多少?

题解

考虑一下最小生成树,实际上在这道题可以理解为0边的联通块一共有多少个,如果有k个,那么答案就是k-1.

然后代码需要优化一下,我们维护两个集合,一个集合是当前已经聚合成连痛块的点,和一堆还在外面没有讨论过的点。

考虑均摊,我们每个点只会访问一次,每条边也只会访问一次,那么复杂度就是(n+m)logn <---- set的复杂度。但这个vis数组的复杂度有点迷,不太会算。。。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 100005;
set<int>g[maxn];
set<int>vis;
int u[maxn];
void dfs(int x){
	u[x]=1;
	vis.erase(x);
	vector<int>ret;
	for(int p:vis){
		if(!g[x].count(p)){
			ret.push_back(p);
		}
	}
	for(int t:ret){
		vis.erase(t);
	}
	for(int t:ret){
		u[t]=1;
		dfs(t);
	}
}
int main(){
	int n,m;
	cin>>n>>m;
	for(int i=0;i<m;i++){
		int x,y;
		cin>>x>>y;
		x--,y--;
		g[x].insert(y);
		g[y].insert(x);
	}
	for(int i=0;i<n;i++){
		vis.insert(i);
	}
	int ans = 0;
	for(int i=0;i<n;i++){
		if(!u[i]){
			ans++;
			dfs(i);
		}
	}
	cout<<ans-1<<endl;
}
posted @ 2019-11-07 17:08  qscqesze  阅读(394)  评论(0编辑  收藏