5-1 File Transfer (30分)

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本文介绍了一个网络文件传输模拟程序的设计与实现。该程序通过输入连接和检查指令来判断网络中任意两台计算机间是否可以进行文件传输，并最终评估整个网络的连通性。

5-1 File Transfer (30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 $N$ ( $2≤N≤1042\le N\le 10^4$ ), 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.

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int main(void)
{
	int n;
	scanf("%d",&n);
	int a[n+1];
	for(int i=0; i<n+1; i++)
	{
		a[i]=-1;
	}
	char choice;
	int x,y,sum,k1,k2,number,k;
	while((choice=getchar())!='S')
	{
		scanf("%d %d",&x,&y);
		if(choice=='C')
		{
			k1=a[x];
			while(k1!=y&&k1>=0)
			{
				k1=a[k1];
				if(k1==y)
				{
					printf("yes\n");
					break;
				}
				if(k1==a[x])
				break;
			}
			k2=a[y];
			while(k2!=x&&k2>=0)
			{
				k2=a[k2];
				if(k2==x)
				{
					printf("yes\n");
					break;
				}
				if(k2==a[y])
				break;
			}
			if(k1!=y&&k2!=x)
			{
				printf("no\n");
			}
		}
		else if(choice=='I')
		{
			k=x;
			while(a[x]>0)
			{
				x=a[x];
				if(x==k)
				break;
			}
			a[x]=y;
		}
	}
	sum=0;
	for(int i=1; i<n+1; i++)
	{
		if(a[i]<0)
		{
			sum++;
		}
	}
	if(sum==1)
	{
		printf("The network is connected.\n");
	}
	else
	{
		printf("There are %d components.\n",sum);
	}
}