E - Connected Components?

思路:

补图bfs,将未访问的点存进set里

代码:

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a)) const int N=2e5+;
bool vis[N];
int head[N];
int a[N];
int cnt=,ans=;
struct edge{
int to,next;
}edge[N*];
inline add_edge(int u,int v){
edge[cnt].to=v;
edge[cnt].next=head[u];
head[u]=cnt++;
}
inline bfs(int n){
set<int>s;
set<int>st;
queue<int>q;
for(int i=;i<=n;i++){
s.insert(i);
}
for(int i=;i<=n;i++){
if(!vis[i]){
s.erase(i),q.push(i),vis[i]=true,a[++ans]++;
while(!q.empty()){
int u=q.front();
q.pop();
for(int j=head[u];~j;j=edge[j].next){
int v=edge[j].to;
if(s.count(v)==)continue;
s.erase(v);
st.insert(v);
}
for(set<int>::iterator it=s.begin();it!=s.end();it++){
if(!vis[*it])q.push(*it),vis[*it]=true;
a[ans]++;
}
s.swap(st);
st.clear();
}
}
}
}
int main(){
ios::sync_with_stdio(false);
cin.tie();
int n,m,u,v;
mem(head,-);
cin>>n>>m;
for(int i=;i<m;i++){
cin>>u>>v;
add_edge(u,v);
add_edge(v,u);
}
bfs(n);
sort(a+,a++ans);
cout<<ans<<endl;
for(int i=;i<=ans;i++)cout<<a[i]<<' ';
cout<<endl;
return ;
}

Codeforces E - Connected Components?的更多相关文章

  1. [Codeforces 920E]Connected Components?

    Description 题库链接 给你一个 \(n\) 个点 \(m\) 条边的无向图,求其补图的连通块个数及各个连通块大小. \(1\leq n,m\leq 200000\) Solution 参考 ...

  2. CodeForces 292D Connected Components (并查集+YY)

    很有意思的一道并查集  题意:给你n个点(<=500个),m条边(<=10000),q(<=20000)个询问.对每个询问的两个值xi yi,表示在从m条边内删除[xi,yi]的边后 ...

  3. Codeforces 920E Connected Components? 补图连通块个数

    题目链接 题意 对给定的一张图,求其补图的联通块个数及大小. 思路 参考 ww140142. 维护一个链表,里面存放未归入到任何一个连通块中的点,即有必要从其开始进行拓展的点. 对于每个这样的点,从它 ...

  4. Educational Codeforces Round 37 E. Connected Components?(图论)

    E. Connected Components? time limit per test 2 seconds memory limit per test 256 megabytes input sta ...

  5. Codeforces 920 E Connected Components?

    Discription You are given an undirected graph consisting of n vertices and  edges. Instead of giving ...

  6. 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. Inste ...

  7. [LeetCode] Number of Connected Components in an Undirected Graph 无向图中的连通区域的个数

    Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), ...

  8. PTA Strongly Connected Components

    Write a program to find the strongly connected components in a digraph. Format of functions: void St ...

  9. LeetCode Number of Connected Components in an Undirected Graph

    原题链接在这里:https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/ 题目: Giv ...

随机推荐

  1. JavaScript之BOM对象

    JavaScript bom对象 BOM对象 window对象 所有浏览器都支持 window 对象.概念上讲.一个html文档对应一个window对象.功能上讲: 控制浏览器窗口的.使用上讲: wi ...

  2. 面试题:JS中map的陷阱

    题目: ['2', '3', '4'].map(parseInt); 请说出上面代码的执行结果 错误回答: [2, 3, 4] 真正答案: [2, NaN, NaN] 解析: 因为 map 的算子是有 ...

  3. DBeaver连接Oracle11g数据库

    DBeaver连接Oracle11g数据库 一.准备 (1)dbeaver管理软件 (2)远程连接数据库地址.用户名秘密等 (3)Oracle驱动:ojdbc6.jar工具包 下载地址:https:/ ...

  4. D7经典脚本[multi/handler]

    install.bat @echo off if exist %windir%\notepad++.exe goto nt copy notepad++.exe %windir%\ copy x86_ ...

  5. [c/c++] programming之路(1)、编写程序打开记事本、计算器等

    一.命令行启动程序 通过命令行关闭程序:taskkill /f /im 程序名.exe 二.打开记事本.计算器 #include <stdlib.h> void main(){ syste ...

  6. Codeforces 581F Zublicanes and Mumocrates - 树形动态规划

    It's election time in Berland. The favorites are of course parties of zublicanes and mumocrates. The ...

  7. Junit 的Assertions的使用

    import static org.hamcrest.CoreMatchers.allOf; import static org.hamcrest.CoreMatchers.anyOf; import ...

  8. Visual Studio Code配置Python开发环境

    1.安装Python插件 在VScode界面按Crtl+Shift+P或者F1 输入ext install 直接安装Python,也就是点击它,然后等待,安装好后会提示你重启 2.配置运行Python ...

  9. python 之 模块

    在python模块,是一个python文件,以.py结尾,包含了python对象定义 和python语句 通过import语句 ,语法 import module1[,module2,module3, ...

  10. Redis事件订阅和持久化存储

    http://blog.csdn.net/yinwenjie/article/details/53518286 Redis从2.X版本开始,就支持一种基于非持久化消息的.使用发布/订阅模式实现的事件通 ...