05-树8 File Transfer (25 分)
We have a network of computers and a list of bi-directional connections. Each of these connections allows a file transfer from one computer to another. Is it possible to send a file from any computer on the network to any other?
Input Specification:
Each input file contains one test case. For each test case, the first line contains N (2), the total number of computers in a network. Each computer in the network is then represented by a positive integer between 1 and N. Then in the following lines, the input is given in the format:
I c1 c2
where I stands for inputting a connection between c1 and c2; or
C c1 c2
where C stands for checking if it is possible to transfer files between c1 and c2; or
S
where S stands for stopping this case.
Output Specification:
For each C case, print in one line the word "yes" or "no" if it is possible or impossible to transfer files between c1 and c2, respectively. At the end of each case, print in one line "The network is connected." if there is a path between any pair of computers; or "There are k components." where k is the number of connected components in this network.
Sample Input 1:
5
C 3 2
I 3 2
C 1 5
I 4 5
I 2 4
C 3 5
S
Sample Output 1:
no
no
yes
There are 2 components.
Sample Input 2:
5
C 3 2
I 3 2
C 1 5
I 4 5
I 2 4
C 3 5
I 1 3
C 1 5
S
Sample Output 2:
no
no
yes
yes
The network is connected.
#include<cstdio>
const int maxn = ; void Initialization( int n);
void Input_connection();
int FindFather(int x);
void Check_connection();
void Check_network(int n);
int arr[maxn]; int main()
{
int n;
char in; scanf("%d",&n); Initialization(n); do
{
getchar();
scanf("%c",&in);
switch(in)
{ case 'I':
{
Input_connection();
break;
}
case 'C':
{
Check_connection();
break;
}
case 'S':
{
Check_network(n);
break;
}
}
}while (in != 'S');
return ; } void Initialization( int n)
{
for (int i = ; i <= n; i++)
{
arr[i] = i;
}
} void Input_connection()
{
int u,v;
scanf("%d %d",&u,&v);
int faA = FindFather(u);
int faB = FindFather(v);
if (faA != faB)
{
arr[faA] = faB;
}
} int FindFather(int x)
{
if (arr[x] == x)
{
return x;
}
else
{
return arr[x] = FindFather(arr[x]);
}
} void Check_connection()
{
int u,v;
scanf("%d %d",&u,&v);
int faA = FindFather(u);
int faB = FindFather(v);
if (faA != faB)
{
printf("no\n");
}
else
{
printf("yes\n");
}
} void Check_network(int n)
{
int cnt = ;
for (int i = ; i <= n; i++)
{
if (arr[i] == i)
{
cnt++;
}
} if ( == cnt)
{
printf("The network is connected.");
}
else
{
printf("There are %d components.",cnt);
}
}
05-树8 File Transfer (25 分)的更多相关文章
- PTA 05-树8 File Transfer (25分)
题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/670 5-8 File Transfer (25分) We have a netwo ...
- PAT 5-8 File Transfer (25分)
We have a network of computers and a list of bi-directional connections. Each of these connections a ...
- L2-006 树的遍历 (25 分) (根据后序遍历与中序遍历建二叉树)
题目链接:https://pintia.cn/problem-sets/994805046380707840/problems/994805069361299456 L2-006 树的遍历 (25 分 ...
- pat04-树5. File Transfer (25)
04-树5. File Transfer (25) 时间限制 150 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 CHEN, Yue We have ...
- L2-006 树的遍历 (25 分)
链接:https://pintia.cn/problem-sets/994805046380707840/problems/994805069361299456 题目: 给定一棵二叉树的后序遍历和中序 ...
- 7-3 树的同构(25 分) JAVA
给定两棵树T1和T2.如果T1可以通过若干次左右孩子互换就变成T2,则我们称两棵树是“同构”的. 例如图1给出的两棵树就是同构的,因为我们把其中一棵树的结点A.B.G的左右孩子互换后,就得到另外一棵树 ...
- 05-树8 File Transfer (25 分)
We have a network of computers and a list of bi-directional connections. Each of these connections a ...
- PAT 甲级 1021 Deepest Root (25 分)(bfs求树高,又可能存在part数part>2的情况)
1021 Deepest Root (25 分) A graph which is connected and acyclic can be considered a tree. The heig ...
- 05-树8 File Transfer(25 point(s)) 【并查集】
05-树8 File Transfer(25 point(s)) We have a network of computers and a list of bi-directional connect ...
随机推荐
- 《JavaEE开发的颠覆者——Spring Boot实战》是一本好书
这本书的风格非常好.每一节都是先点明这一块知识的要点,随后就手把手的做出一个最简明.但有能体现核心的实例(大多只有几个Class) 这样的书用来熟悉一门框架,实在是再好不过.
- 7.JavaScript-Promise的并行和串行
Promise 并行 Promise.all是所有的Promise执行完毕后(reject|resolve)返回一个Promise对象. 最近在开发一个项目中,需要等接口拿到全部数据后刷新页面,取消l ...
- Java 控制流程 之 循环语句
循环:循环语句可以在满足循环条件的情况下,反复执行某一段代码,这段被重复执行的代码被称为循环体语句,当反复 执行这个循环体时,需要在合适的时候把循环判断条件修改为false,从而结束循环,否则循环将一 ...
- Android实现二维码扫描功能
1.效果预览 先上图展示效果(模拟器没有摄像头,录出来效果不好,将就看) 2.集成步骤 1.拷贝本项目demo中的com.google.zxing5个包引入到自己的项目中. 2.拷贝本项目demo中的 ...
- git https解决免ssL和保存密码
1.打开windows的git bash set GIT_SSL_NO_VERIFY=true git clonegit config --global http.sslVerify false 2 ...
- 配置集成测试环境 phpstudy
phpStudy是一个PHP调试环境的程序集成包,该程序包集成最新的Apache+PHP+MySQL+phpMyAdmin+ZendOptimizer,一次性安装,无须配置即可使用,是非常方便.好用的 ...
- Linux搭建MySQL主从
实现目标 搭建两台MySQL服务器(一主一从),一台作为主服务器,一台作为从服务器,主服务器进行写操作,从服务器进行读操作. 工作流程概述 主服务器: 开启二进制日志 配置唯一的server-id 获 ...
- python json dumps与loads
json.dumps() 是将一个Python数据结构转换为一个JSON编码的字符串 json.loads() 是将一个JSON编码的字符串转换为一个Python数据结构 一般要求当要字符串通过l ...
- Nginx 反向代理与负载均衡的配置
已经很久没有写博了,因为最近学车加上各种问题一直没时间, 今天刚好想起有好多的东西还没来得及记录.回到正题: Nginx是一个非常强大的web轻量级服务器,许多大厂也用Nginx进行负载均衡和反向代理 ...
- manjaro手动安装Redis
以前都是用的Windows系统,最近有被win10搞得有点烦,就入了manjaro的坑,windows下部分软件在manjaro安装记录,留个记录. 我的系统信息 下面开始正式干活. 一.准备步骤 下 ...