PTA 05-树8 File Transfer (25分)

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原文链接：http://www.cnblogs.com/gk2017/p/7140845.html

本文介绍了一个网络连接检测算法实现，并通过实例演示了如何判断任意两台计算机间是否能传输文件。该算法使用并查集数据结构来维护网络中各计算机之间的连接状态。

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题目地址

https://pta.patest.cn/pta/test/16/exam/4/question/670

5-8 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 $(2\le N\le 10^42≤N≤104), the total number of computers in a network. Each computer in the network is then represented by a positive integer between 1 and NN. Then in the following lines, the input is given in the format:$

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.


这题是考并查集

/*
运行详情
评测结果
时间	结果	得分	题目	编译器	用时（ms）	内存（MB）	用户
2017-06-29 18:12	部分正确	24	5-8	gcc	27	1	
测试点结果
测试点	结果	得分/满分	用时（ms）	内存（MB）
测试点1	答案正确	7/7	18	1
测试点2	答案正确	7/7	1	1
测试点3	答案正确	1/1	2	1
测试点4	答案正确	1/1	2	1
测试点5	答案正确	4/4	14	1
测试点6	答案正确	4/4	15	1
测试点7	答案错误	0/1	27	1

没有ac，不知道最后一个测试点用了什么数据 
*/

#include<stdio.h>
#define MAXLEN 15000
#define TOPLEVEL -1
int gComputers[MAXLEN];
void InitComputers(int n)
{
	int i;
	for(i=1;i<=n;i++)
	{
		gComputers[i]=TOPLEVEL;
	}
}
void swap(int *a,int *b)
{
	int temp;
	temp=*a;
	*a=*b;
	*b=temp;	
}
void Connect(int a,int b)
{
/*	if(gComputers[a]==TOPLEVEL)
	{
		gComputers[b]=a;
		return;
	}
*/	
	int temp;
	temp=a;
	while(gComputers[temp]!=TOPLEVEL)
	{
		temp=gComputers[temp];
	}
//		gComputers[a]=temp;
		gComputers[b]=temp;
		return;
}

/*
int GetTop(int a)
{
	
	int temp=a;
	if(gComputers[a]==TOPLEVEL)
		return a;

		while(gComputers[temp]!=TOPLEVEL)
			temp=gComputers[temp];
		gComputers[a]=temp; //查找完了合并父节点； 
		return temp;
		
}
*/
int GetTop(int k)  
{  
    if(gComputers[k]<0)  
    {  
        return k;  
    }  
    return gComputers[k] = GetTop(gComputers[k]);  
}  

void Check(int a,int b)
{
	if(GetTop(a)==GetTop(b))
	{
		printf("yes\n");
	}
	else
	{
		printf("no\n");
	}
}

void ShowSummry(int n)
{
	int i,count=0;
	for(i=1;i<=n;i++)//此处扫描应从1号开始，电脑从1开始编号，0号没操作始终是-1； 
	{
		if(gComputers[i]==TOPLEVEL)
		{
			count++;
		}
	}
	if(count==1)
	{
		printf("The network is connected.");
	}
	else
		printf("There are %d components.",count);
}

void OutputStatus(int n)
{
	int i;
	for(i=0;i<=n;i++)
	{
		printf("--%d:%d\n",i,gComputers[i]);
	}
}

int main()
{
	int N;
	int a,b;
	char command;
	scanf("%d",&N);
	InitComputers(N);
	while(1)
	{
		scanf("%c",&command);
		if(command=='S')
		{
			ShowSummry(N);
//			OutputStatus(N);
			break;
		}
			
		scanf("%d %d",&a,&b);
		if(command=='I')
			Connect(a,b);
		if(command=='C')
			Check(a,b);
	}
}

不知道最后一个测试点什么情况，mooc开课之外的题库里面不带提示，找了一张别人做题的截图，发现最后一个点是卡不压缩的，容易超时

问题是我有一直合并路线，而且时间也不长。不知道遇上什么问题了。