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
- 怎样调节Eclipse中的字体大小?
window->perference->appearance->colors and font->text font edit
- SUSE12SP3-Samba配置
简介 samba官网:https://www.samba.org/ 维基百科: https://zh.wikipedia.org/wiki/Samba Samba,是种用来让UNIX系列的操作系统与微 ...
- 前端有用的CSS属性和JS方法
1.CSS属性: 透明属性(值越大越不透明): IE:filter:alpha(opacity:30) Google:opacity:0.3 层次属性(值大的会在上面): z-index:100 2. ...
- 学习笔记之Django
The Web framework for perfectionists with deadlines | Django https://www.djangoproject.com/ Django m ...
- Java 之 Scanner 类
一.Scanner 类 Scanner 是一个可以解析基本类型和字符串的简单文本扫描器. Demo: Scanner sc = new Scanner(System.in); int i = sc.n ...
- python爬虫爬取天气数据并图形化显示
前言 使用python进行网页数据的爬取现在已经很常见了,而对天气数据的爬取更是入门级的新手操作,很多人学习爬虫都从天气开始,本文便是介绍了从中国天气网爬取天气数据,能够实现输入想要查询的城市,返回该 ...
- C++使用通配符查找文件(FindFirstFile)
调用 FindFirstFile 和 FindNextFile 可搜索某个目录下的相应文件. BOOL SearchFilesByWildcard(WCHAR *wildcardPath) { HAN ...
- C语言 dlopen dlsym
C语言加载动态库 头文件:#include<dlfcn.h> void * dlopen(const char* pathName, int mode); 返回值 handle void ...
- 2019年牛客多校第二场 F题Partition problem 爆搜
题目链接 传送门 题意 总共有\(2n\)个人,任意两个人之间会有一个竞争值\(w_{ij}\),现在要你将其平分成两堆,使得\(\sum\limits_{i=1,i\in\mathbb{A}}^{n ...
- ImportError: No module named wx
先检查下Python中是否有此模块 $ python >>> import wx ImportError: No module named wx wx模块的确没有安装. 要安装wx ...