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 $N$ (2\le N\le 10^42≤N≤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

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.

思路：本題的關鍵點是按秩歸併和路徑壓縮，關於這兩點在MOOC上老師有很詳細的解釋。

#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;

int S[10001];
int N;

int Find(int S[],int i)
{
	if(S[i] < 0)
	{
		return i;
	}
	else
	return  S[i] = Find(S,S[i]);/*不斷的找父結點，一直找到根結點，然後再把每個父結點指向根結點*/
}

void check(int S[])
{
	int root1,root2;
	int a, b;
	scanf("%d %d",&a,&b);
	root1 = Find(S,a-1);
	root2 = Find(S,b-1);
	
	if(root1 == root2)
	{
		printf("yes\n");
	}
	else
	printf("no\n");
}

void Union(int S[],int root1,int root2)
{
	if(S[root2] < S[root1])
	{
		S[root2] += S[root1];
		S[root1] = root2;
	}
	else
	{
		S[root1] += S[root2];
		S[root2] = root1;
	}
}

void Insert(int S[])
{
	int root1,root2;
	int a, b;
	scanf("%d %d",&a,&b);
	root1 = Find(S,a-1);
	root2 = Find(S,b-1);
	if(root1 != root2)
	{
		Union(S,root1,root2);
	}
}

int main(void)
{
	char ch;
	int count = 0;
	scanf("%d",&N);
	for(int i = 0; i < N; i++)/*初始化令每個結點都是根結點*/
	{
		S[i] = -1;
	}
	while(1)
	{
		scanf("%s",&ch);
		
		if(ch == 'C')
		{
			check(S);
		}
		else if(ch == 'I')
		{
			Insert(S);
		}
		else if(ch == 'S')
		{
			break;
		}
	}
	
	for(int i = 0; i < N; i++)
	{
		if(S[i] < 0)count++;
	}
	if(count == 1)
	{
		printf("The network is connected.");
	}
	else
	{
		printf("There are %d components.",count);
	}
	return 0;
}

在之前的Find函數中我用的是循環來返回根結點，結果犯了一個錯誤。

int Find(int S[],int i)

{

for(;S[i] > 0; i = S[i]);

return i;

}

for循環後的那個分號我沒加，結果導致i = S[i]這條語句在for循環判斷後（不滿足S[i] > 0）,仍然被執行。