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

Educational Codeforces Round 37 (Rated for Div. 2) E. Connected Components? 图论

E. Connected Components?

You are given an undirected graph consisting of n vertices and edges. Instead of giving you the edges that exist in the graph, we give you m unordered pairs (x, y) such that there is no edge between x and y, and if some pair of vertices is not listed in the input, then there is an edge between these vertices.

You have to find the number of connected components in the graph and the size of each component. A connected component is a set of vertices X such that for every two vertices from this set there exists at least one path in the graph connecting these vertices, but adding any other vertex to X violates this rule.

Input

The first line contains two integers n and m (1 ≤ n ≤ 200000, ).

Then m lines follow, each containing a pair of integers x and y (1 ≤ x, y ≤ n, x ≠ y) denoting that there is no edge between x and y. Each pair is listed at most once; (x, y) and (y, x) are considered the same (so they are never listed in the same test). If some pair of vertices is not listed in the input, then there exists an edge between those vertices.

Output

Firstly print k — the number of connected components in this graph.

Then print k integers — the sizes of components. You should output these integers in non-descending order.

Example

input
5 5
1 2
3 4
3 2
4 2
2 5
output
2
1 4

题意

给你n个点的完全图,告诉你有m条边是不可连的。问你里面一共有多少个联通块,输出每个块的大小。

题解

https://www.cnblogs.com/qscqesze/p/11813351.html 一摸一样

经验get

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 200005;
int n,m;
set<int>S[maxn];
set<int>vis;
int v[maxn];
int dfs(int x){
	int now = 1;
	vector<int> ret;
	for(int v:vis){
		if(!S[x].count(v))
			ret.push_back(v);
	}
	for(int i=0;i<ret.size();i++){
		vis.erase(ret[i]);
	}
	for(int i=0;i<ret.size();i++){
		v[ret[i]]=1;
		now+=dfs(ret[i]);
	}
	return now;
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++){
		int x,y;cin>>x>>y;
		x--,y--;
		S[x].insert(y);
		S[y].insert(x);
	}
	vector<int> ans;
	for(int i=0;i<n;i++){
		vis.insert(i);
	}
	for(int i=0;i<n;i++){
		if(!v[i]){
			ans.push_back(dfs(i));
		}
	}
	cout<<ans.size()<<endl;
	sort(ans.begin(),ans.end());
	for(int i=0;i<ans.size();i++){
		cout<<ans[i]-1<<" ";
	}
	cout<<endl;
}
posted @ 2019-11-07 18:26  qscqesze  阅读(85)  评论(0编辑  收藏