PTA_2019春_058_ File Transfer

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最新推荐文章于 2022-04-22 22:13:23 发布

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分类专栏：数据结构PTA

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该博客讨论了如何分析一个由N台电脑组成的网络中，通过双向连接进行文件传输的可能性。输入规格定义了每台电脑用整数表示，以及连接、查询和结束的格式。输出规格说明了在给定查询下，文件是否可以在两台电脑间传输，以及网络的连通性状态。示例输入和输出展示了具体的操作和判断结果。

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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.

/*File Transfer */
/*思路：本题应该是涉及集合的问题，即两个电脑连接则属于同一个集合
        根据二者元素是否在同一个集合中判断是否相连
		根据集合中有几个子集合，即有几个组件
		如果所有元素都在一个集合中，则全部相联*/
#include<stdio.h>
#include<stdlib.h>
void connect(int* array, int a, int b);
int find(int* array, int x);
void check(int* array, int a, int b);
int main() {
	int N;/*节点个数*/
	char laji;/*除去上一行的\n*/
	scanf("%d", &N);
	int* array = (int*)malloc((N + 1) * sizeof(int));

	/*初始化*/
	for (int i = 1; i < N + 1; i++) {
		array[i] = -i;
	}
	char a;/*字符*/
	int c1, c2;
	while (1) {
		scanf("%c", &laji);
		scanf("%c", &a);
		if (a == 'I') {  /*用单引号*/
			scanf("%d", &c1);
			scanf("%d", &c2);
			connect(array, c1, c2);
		}
		else if (a == 'C') {
			scanf("%d", &c1);
			scanf("%d", &c2);
			check(array, c1, c2);
		}
		else  break;
	}
	int cnt = 0;/*计数*/
	for (int i = 1; i < N + 1; i++) {  /*cnt为集合内子集合的个数*/
		if (array[i] < 0) cnt++;
	}
	if (cnt == 1) printf("The network is connected.");
	else   printf("There are %d components.",cnt);

	return 0;
}
void connect(int* array, int a, int b) {   /*将较小元素作为根*/
	a = find(array, a); /*找到它的根节点*/
	b = find(array, b);
	if (a < b) {
		array[b] = a;
	}
	else {
		array[a] = b;
	}
}
int find(int* array, int x) {   /*路径压缩*/
	if(array[x]<0){
		return x;
	}
	else{
		return array[x]= find(array,array[x]);
	}
}
/*不压缩 
int find(int* array, int x) {
	for (; array[x] > 0; x = array[x]);
	return x;
}
*/
void check(int* array, int a, int b) {
	a = find(array, a); /*找到它的根节点*/
	b = find(array, b);
	if (a == b) printf("yes\n");
	else printf("no\n");
}