PTA 7-1 File Transfer

最新推荐文章于 2022-10-28 19:47:41 发布

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最新推荐文章于 2022-10-28 19:47:41 发布

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本文介绍了一个利用并查集算法解决网络中计算机间文件传输连通性问题的方法。通过输入网络中计算机的数量及连接关系，算法能够判断任意两台计算机之间是否可以进行文件传输，并最终确定整个网络是否完全连通。

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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.

解题思路：

此问题可用并查集求解。

查找两个电脑是否相连，只需访问他们的根是否相同即可。

所求结果若是电脑全部相连，即所有的数字对应的根相同，输出 "The network is connected." ；

否则就输出有几个连接的集合，这里我用map函数来统计。

代码如下：

#include<iostream>
#include<map>
#include<algorithm>
using namespace std;
typedef struct node
{
	int data;
	int rank;
	int parent;
}UFSTree;

void MAKE_SET(UFSTree t[], int n)
{
	int i;
	for (i = 0; i <= n; i++)
	{
		t[i].data = i;
		t[i].rank = 0;
		t[i].parent = i;
	}
}

int FIND_SET(UFSTree t[], int x)
{
	if (x != t[x].parent)
		return(FIND_SET(t, t[x].parent));
	else
		return x;
}

void UNION(UFSTree t[], int x, int y)
{
	x = FIND_SET(t, x);
	y = FIND_SET(t, y);
	if (t[x].rank > t[y].rank)
		t[y].parent = x;
	else
	{
		t[x].parent = y;
		if (t[x].rank == t[y].rank)
			t[y].rank++;
	}
}

int main()
{
	int N;
	cin >> N;
	UFSTree P[10001];
	MAKE_SET(P, N);

	while (1)
	{
		char c;
		cin >> c;

		if (c == 'S')
			break;

		int a, b;
		cin >> a >> b;
		if (c == 'I')
			UNION(P, a, b);
		else if (FIND_SET(P, a) == FIND_SET(P, b))
			cout << "yes" << endl;
		else
			cout << "no" << endl;
	}

	map<int, int>mp;
	int sum = 0;
	for (int i = 1; i <= N; i++)
	{
		if (!mp.count(FIND_SET(P, i)))
		{
			mp[FIND_SET(P, i)] = 1;
			sum++;
		}
	}
	if (sum == 1)
		cout << "The network is connected." << endl;
	else
		cout << "There are "<<sum<<" components." << endl;
	return 0;
}