04-2. File Transfer (PAT) - 集合问题

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

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." wherek 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.

题意：根据给出的数字，建立这些数字的集合关系解题思路：使用数组存储，数组下标对应给出的元素，数组元素值对应该下标值的parent]如数组a[3] = 8表示，元素3的parent为8注意：对于大规模的集合，可能产生某个集合层级很多的情况，可在进行查找操作时减少集合的层级。

#include <iostream>
#include <string>
using namespace std;

#define MaxNum 10001

int Find( int arrSet[], int X );
void ConnectSet( int arrSet[], int c1, int c2 );
string CheckConnection( int arrSet[], int c1, int c2 );
int CountComponentNum( int arrSet[], int num );

//使用数组保存集合元素，数组下标对应元素值，数组元素对应parent值

int main()
{
    int nodeNum;
    int *arrSet;
    cin >> nodeNum;
    nodeNum += 1;    //由于数组下标0用于标记根结点，因此使用数组下标从1开始
    arrSet = new int[ nodeNum ];    //parent == 0 -> root, parent == -1 -> NULL
    int i;
    for ( i = 0; i < nodeNum; i++ )
    {
        arrSet[i] = -1;
    }

    int c1;
    int c2;
    string operationName;
    string checkResult = "";
    int componentNum;
    while( true )
    {
        cin >> operationName;
        if ( operationName == "S" )
        {
            break;
        }
        cin >> c1 >> c2;
        if ( operationName == "I" )
        {
            ConnectSet( arrSet, c1, c2 );
        }
        else if ( operationName == "C" )
        {
            checkResult += CheckConnection( arrSet, c1, c2 );
        }
    }
    cout << checkResult;
    componentNum = CountComponentNum( arrSet, nodeNum );
    if ( componentNum == 1 )
    {
        cout << "The network is connected." << endl;
    }
    else
    {
        cout << "There are " << componentNum << " components." << endl;
    }
}

int Find( int arrSet[], int X )    //在查找过程中，减少树的高度
{
    if ( X > MaxNum || arrSet[X] == -1 )
    {
        return -1;
    }
    int temp = X;
    int exchange;
    while( arrSet[X] > 0 )
    {
        X = arrSet[X];
    }
    while ( arrSet[arrSet[temp]] > 0 )    //将搜索路径中的所有元素都直接指向根结点
    {
        exchange = temp;
        temp = arrSet[temp];
        arrSet[exchange] = X;
    }
    return X;
}

void ConnectSet( int arrSet[], int c1, int c2 )
{
    int root1, root2;
    if ( arrSet[c1] == -1 )
    {
        arrSet[c1] = 0;
        root1 = c1;
    }
    else
    {
        root1 = Find( arrSet, c1 );
    }
    if ( arrSet[c2] == -1 )
    {
        arrSet[c2] = 0;
        root2 = c2;
    }
    else
    {
        root2 = Find( arrSet, c2 );
    }
    if ( root1 != root2 )
    {
        arrSet[root2] = root1;
    }
}

string CheckConnection( int arrSet[], int c1, int c2 )
{
    int root1, root2;
    root1 = Find( arrSet, c1 );
    root2 = Find( arrSet, c2 );
    if ( root1 == root2 && root1 != -1 && root2 != -1 )
    {
        return "yes\n";
    }
    else
    {
        return "no\n";
    }
}

//计算数组中的集合个数
int CountComponentNum( int arrSet[], int num )
{
    int i;
    int sum = 0;
    int nodeNum = 0;
    for( i = 0; i < num; i++ )
    {
        if ( arrSet[i] >= 0 )    //统计所有结点数量
        {
            nodeNum++;
        }
        if ( arrSet[i] == 0 )    //根结点的数量
        {
            sum++;
        }
    }
    return sum + num - 1 - nodeNum;    //给出的结点数中，可能存在未包含进任何集合的结点                                       //这里-1是由于在一开始将nodeNum加了1,这个处理需要改进
}