05-树8 File Transfer

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 NN (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 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

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.

题目大意：判断两台电脑是否连通，并判断电脑间是否相互连通，否则输出最大连通数目

思路：可以通过集合来解决这道题。利用判断两台电脑是否在同一集合里，若是则连通，否则不连通。

#include<iostream>
using namespace std;

int *s;
int FindRoot(int x)
{
if (s[x] == x)
return x;
else
s[x] = FindRoot(s[x]);
}

void Union(int a, int b)
{
int root1, root2;
root1 = FindRoot(a);      //找根
root2 = FindRoot(b);
if (root1 != root2){
if (s[root1]<s[root2])
s[root2] = root1;
else
s[root1] = root2;
}
}

int main()
{
int i, N, num1, num2, count = 0;
char ch;
cin >> N;
s = new int[N + 1];
for (i = 0; i <= N; i++){
s[i] = i;
}
while (1){
cin >> ch;
if (ch == 'S')
break;
cin >> num1 >> num2;
if (ch == 'I')
Union(num1, num2);
if (ch == 'C'){
if (FindRoot(num1) == FindRoot(num2))
cout << "yes" << endl;
else
cout << "no" << endl;
}
}
for (i = 1; i <= N; i++){
if (s[i] == i)
count++;
}
if (count == 1)
cout << "The network is connected." << endl;
else
cout << "There are " << count << " components." << endl;
delete[] s;
return 0;
}