sicily 4378 connected components in undirected graph
题意:求图中的连通块数,注意孤立的算自连通!
例如:6个顶点3条路径,其中路径为:1->2 4->5 1->3
那么有(1-2&&1->3) + (4->5) + (6) 共3个连通块!
解法:对每个节点宽搜!
#include<iostream>
#include<memory.h>
#include<queue> using namespace std; bool roads[][];
bool visited[];
int N,M; int main(){ cin >>N >>M;
memset(roads,,sizeof(roads));
memset(visited,false,sizeof(visited));
int from,dest;
for(int i=; i<=M; i++){
cin >> from >> dest;
roads[from][dest] = true;
roads[dest][from] = true;
} queue<int> check;
int num = ;
int cnt = ;
int i;
//breadth-frist search
while(num != N){
for( i=; i<=N;i++){
if(visited[i]== false){
check.push(i);
visited[i]= true;
num++;
cnt++;
break;
}
}
while(!check.empty()){
i = check.front();
for(int j = ; j<=N;j++){
if(roads[i][j] == true && visited[j] == false){
check.push(j);
visited[j] = true;
num++;
}
}
// erase the front node
check.pop();
}
}
cout << cnt <<endl;
return ;
}
sicily 4378 connected components in undirected graph的更多相关文章
- Connected Component in Undirected Graph
Description Find connected component in undirected graph. Each node in the graph contains a label an ...
- Sicily connect components in undirected graph
题目介绍: 输入一个简单无向图,求出图中连通块的数目. Input 输入的第一行包含两个整数n和m,n是图的顶点数,m是边数.1<=n<=1000,0<=m<=10000. 以 ...
- [SOJ] connect components in undirected graph
题目描述: 输入一个简单无向图,求出图中连通块的数目 输入: 输入的第一行包含两个整数n和m,n是图的顶点数,m是边数.1<=n<=1000,0<=m<=10000. 以下m行 ...
- 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 ...
- Codeforces 920 E Connected Components?
Discription You are given an undirected graph consisting of n vertices and edges. Instead of giving ...
- 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 ...
- CodeForces242D:Connected Components (不错的并查集)
We already know of the large corporation where Polycarpus works as a system administrator. The compu ...
- D. Connected Components Croc Champ 2013 - Round 1 (并查集+技巧)
292D - Connected Components D. Connected Components time limit per test 2 seconds memory limit per t ...
- [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), ...
随机推荐
- git查看某个文件的修改历史
<转自 http://www.cnblogs.com/flyme/archive/2011/11/28/2265899.html> 有时候在比对代码时,看到某些改动,但不清楚这个改动的作者 ...
- 零基础学习云计算及大数据DBA集群架构师【Linux Bash Shell编程及系统自动化2015年1月21日周四】
lvy老师教项目课程,以及代课了shell部分课程,大家都觉得这位老师不行,上课时做的操作很多都是错误的,觉得她基础不好.而且,她不能解释原因,学生问为什么,她不知道.崩溃啊.向xx培训机构反应后,说 ...
- HDU 1199 - Color the Ball 离散化
[题意]现在有几个球排成一排,编号从1开始,开始时所有球为黑色,现在有n(<=2000)次操作,每次操作将l[i]至r[i](均在int范围)的球凃成颜色c[i](黑色'b'或白色'w'),然后 ...
- 单例模式 与lock用法
在之前没用lock之前:如果我实现单例模式:直接就是下面的代码: public class Singleton { private static Singleton instanc ...
- django安装
见 http://jingyan.baidu.com/article/466506580e7d29f549e5f8b6.html 下载安装python下载解压django cmd进入django目录, ...
- php中curl、fsockopen的应用
最近要用到通过post上传文件,网上盛传的有curl的post提交和fsockopen,其中curl最简单,于是从最简单的说起. 这是简单的将一个变量post到另外一个页面 $url = ''; $d ...
- 如果设置http.get超时控制
var timeout_wrapper = function (req) { return function () { // do some logging, cleaning, etc. depen ...
- Oracle数据库之PL/SQL流程控制语句
Oracle数据库之PL/SQL流程控制语句 在任何计算机编程语言(如C,Java,C#等)都有各种流程控制语句,同样,在PL/SQL中也存在这样的流程控制结构. 几种常见的流程控制结构: 一.条件结 ...
- java Socket 长连接 心跳包 客户端 信息收发 demo
今天写了个socket的测试小程序,代码如下 import java.io.IOException; import java.io.InputStream; import java.io.Output ...
- 百度touch的手势框架,touch.js
今天,随便搜搜看到了一个新的手势库,也许能让我为现在使用者的hammer.js的手势库带来的烦恼而消除. 它是百度团队开发的,现在由百度云Clouda进行维护. 官网 http://touch.c ...