File Transfer

最新推荐文章于 2022-05-04 14:51:23 发布

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最新推荐文章于 2022-05-04 14:51:23 发布

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分类专栏： mooc数据结构文章标签：数据结构 namespace 编程

本文链接：https://blog.youkuaiyun.com/Phenixfate/article/details/42294345

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该博客探讨了一个网络中文件传输的可能性问题，通过处理输入的计算机连接和查询，判断任意两台计算机之间是否存在传输路径。解决方案涉及并查集的数据结构，可以使用数组来实现。博主分享了用树形结构数组解决此问题的经验，并提到了一位大神用简洁的数组方法解决此问题的链接。

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先粘个题目不解释：

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<=N<=10⁴), 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

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:

Sample Output 1:

no
no
yes
There are 2 components.

Sample Input 2:

Sample Output 2:

no
no
yes
yes
The network is connected.

解释下题目的意思：

这道题就是一个并查的问题，也就是判断2个数是否属于同一个集合

输入格式：

第一行先输入结点的总数量

接着就是输入需要进行的操作，有三种：C,I,S

C表示检查2个数是不是在一个集合，I 就是将后面的两个数合并到为一个集合，S就是停止了

输出的话：

就是先判断那些进行check的集合，最后还要输出有多少个集合，要注意只有一个是，输出的应为The network is connected.

一开始做这道题时，在想到底改用什么方法来完成，其实这个问题很简单，我们用数学思维的方式一下就能做出来，但是要如何用编程的方式来解决？

我先准备像老师讲的用数组来储存结点的方式来做题，就是采用树的形式，一个结点包括data和 parent两个元素，最后也的确完成了。

后来发现一个大神，直接用数组的方式就解决了，粘上链接:http://www.cnblogs.com/clevercong/p/4192953.html

发现这方法实在是太赞了

直接上代码：

#include <iostream>
#define ElementType int
using namespace std;

int *S;
int Find(ElementType X)
{
	if (S[X] == X)
		return X;
	return S[X] = Find(S[X]);
}

void Union(ElementType X1, ElementType X2)
{
	int Root1, Root2;
	Root1 = Find(X1);
	Root2 = Find(X2);
	if (Root1 != Root2)
	{
		if (S[Root1] < S[Root2])
			S[Root2] = Root1;
		else
			S[Root1] = Root2;
	}
}
 int main()
 {
	 int num;
	 cin >> num;
	 char choose;
	 int c1, c2;
	 S = new int[num + 1];
	 for (int i = 0; i <= num; i++)   // 初始化
		S[i] = i;
	 while (1)
	 {
		 cin >> choose;
		 if (choose == 'S')
			 break;
		 cin >> c1 >> c2;
		 if (choose == 'I')
			 Union(c1, c2);
		 if (choose == 'C')
		 {
			 if (Find(c1) == Find(c2))
				 cout << "yes" << endl;
			 else
				 cout << "no" << endl;
		 }
	 }
	int icount = 0;
	for (int i = 1; i <= num; i++)
		if (S[i] == i)  
			icount++;
	if (icount == 1)
		cout << "The network is connected." << endl;
	else
		cout << "There are " << icount << " components." << endl;
	return 0;
 }

感谢大神的分享！！！