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 $N$ ( $2≤N≤1042\le N\le 10^4$ ), 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 andc2; or

C c1 c2

where C stands for checking if it is possible to transfer files betweenc1 andc2; 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 betweenc1 andc2, 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 arek components." wherek 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.

 这道题考察的是集合的并查运算，同时运用了按秩归并和路径压缩的优化方法。。。同时也让我看到了JAVA和C++的效率。同样的逻辑，java几乎全部超时，而C++一次通过
  这道题的思路是,以数组结构来组织集合（模拟树结构），其中数组的下标值集合元素的值，数组的值是该位置的父节的值（也就是父节点的下标），其中根结点在数组中
的值为负数(-树的高度),

程序初始化的时候,将数组中每个位置的值都设为-1，即再没有连接前，每个节点都是一个独立的集合。

C++AC代码:

#include 
#include 
using namespace std;

int s[10005];
int N;
void input_connect();
void check_connect();
void stop_connect();
int finds(int k);
void unions(int root1,int root2);

int main()
{
    scanf("%d",&N);
    for(int i=0; is[root2])
    {
        s[root1] = root2;
    }
    else
    {
        s[root1] = root2;
        s[root2]--;
    }
}

java超时代码:~~~~~

import java.util.Scanner;
import java.util.Arrays;
public class Main{
	private static int[] s;
	private static Scanner scan = new Scanner(System.in);
	public static void main(String[] args) {
		int N = scan.nextInt();
		s = new int[N];
		Arrays.fill(s,-1);
		char cp;
		do{
			cp = scan.next().charAt(0);
			switch(cp){
				case 'I':input_connect();break;
				case 'C':check_connect();break;
				case 'S':stop_connect(N);break;
			}
		}while(cp!='S');
	}

	public static void input_connect(){
		int a = scan.nextInt();
		int b = scan.nextInt();
		int root1 = find(a-1);
		int root2 = find(b-1);
		if(root1!=root2){
			union(root1,root2);
		}
	}

	public static void check_connect(){
		int a = scan.nextInt();
		int b = scan.nextInt();
		int root1 = find(a-1);
		int root2 = find(b-1);
		if(root1==root2){
			System.out.println("yes");
		}else{
			System.out.println("no");
		}
	}

	public static void stop_connect(int N){
		int cnt = 0;
		for(int i=0;is[root2]){
			s[root1] = root2;
		}else{
			s[root1] = root2;
			s[root2]--;
		}
	}
}

按秩归并:顾名思义，就是在合并两个集合的时候，把高度较小的集合根节点的父节点设为高度较大的父节点，这样可以保证合并后的集合高度最小。路径压缩：就是在查找集合元素的根节点的时候，一旦找到根节点，在回溯的过程中，将其子节点的值全部跟改为根结点的值。这样只在第一次查找的时候比较耗时，但在后续的查找过程中，将会节省大量的时间。