MOOC浙大数据结构 — 05-树9 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 c1and 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 kis 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.

#include 

#define MAXN 10000

int S[MAXN];

void Union( int Root1, int Root2 )
{ /* 这里默认Root1和Root2是不同集合的根结点 */
    /* 保证小集合并入大集合 */
    if ( S[Root2] < S[Root1] ) { /* 如果集合2比较大 */
        S[Root2] += S[Root1];     /* 集合1并入集合2  */
        S[Root1] = Root2;
    }
    else {                         /* 如果集合1比较大 */
        S[Root1] += S[Root2];     /* 集合2并入集合1  */
        S[Root2] = Root1;
    }
}

int Find( int X )
{ /* 默认集合元素全部初始化为-1 */
    if ( S[X] < 0 ) /* 找到集合的根 */
        return X;
    else
        return S[X] = Find( S[X] ); /* 路径压缩 */
}

int main() {

    int n;
    scanf( "%d", &n );
    for( int i = 0; i < n; i++ ) S[i] = -1;
    char ch = getchar();
    while( 1 ) {
        ch = getchar();
        if( ch == 'S' )
            break;
        int a, b;
        scanf( "%d%d", &a, &b );
        int parentA = Find( a );
        int parentB = Find( b );
        if( ch == 'C' ) {
            if( parentA != parentB ) {
                printf( "no\n" );
            }
            else printf( "yes\n" );
        }
        if( ch == 'I' )
            Union( parentA, parentB );
    }
    int count = 0;
    for( int i = 0; i < n; i++ ) {
        //printf( "%d ", S[i] );
        if( S[i] < 0 ) count++;
    }
    //printf( "\n" );
    if( count == 1 ) printf( "The network is connected.\n" );
    else printf( "There are %d components.", count );
    return 0;
}