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
- [Unit Testing] Node testing: Test api Get request
Using mocha: "devDependencies": { "should": "^5.2.0", "supertest& ...
- Genymotion 模拟器 VirtualBox
准备 介绍: 1.Genymotion安卓模拟器其实不是普通的模拟器,严格来说,genymotion是虚拟机,被网传定义为模拟器,加载APP的速度比较快,操作起来也很流畅.2.Genymotion依赖 ...
- C# 窗口传值的方法
方法一: A to B 设置FormB 为 带参数的构造函数 public Form2( object msg) { InitializeComponent(); } 方法二: A to B 定义一 ...
- java反射新的应用
利用java反射动态修改运行中对象的私有final变量,不管有没有get方法获取这个私有final变量. spring aop 本质是cglib,动态代理 可以做很多事情 query.addCrite ...
- (转)js获取url参数值
明天有空编辑下 今天做项目遇到js取得url地址问号后面的参数,找了下面的,用着非常好,项目是选项卡样式的,也就是一点击二级分类,底下的同样名字的背景变红(选项卡倍选中) http://www.cnb ...
- 用C#实现的条形码和二维码编码解码器
本文主要介绍可以在C#中使用的1D/2D编码解码器.条形码的应用已经非常普遍,几乎所有超市里面的商品上面都印有条形码:二维码也开始应用到很多场合,如火车票有二维码识别.网易的首页有二维码图标,用户只需 ...
- PHP MySQLi
PHP MySQLi 简介 PHP MySQLi = PHP MySQL Improved! MySQLi 函数允许您访问 MySQL 数据库服务器. 注释:MySQLi 扩展被设计用于 MySQL ...
- impdp/expdp 总结
impdp/expdp 1.创建DIRECTORY create directory dir_dp as '/tmp'; --建议将DIRECTORY 建在 /tmp 表下面,该目录肯定存 ...
- 记录Linux下安装elasticSearch时遇到的一些错误
记录Linux下安装elasticSearch时遇到的一些错误 http://blog.sina.com.cn/s/blog_c90ce4e001032f7w.html (2016-11-02 22: ...
- python 操作 mysql基础补充
前言 本篇的主要内容为整理mysql的基础内容,分享的同时方便日后查阅,同时结合python的学习整理python操作mysql的方法以及python的ORM. 一.数据库初探 在开始mysql之前先 ...