5-8 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 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:

5C 3 2I 3 2C 1 5I 4 5I 2 4C 3 5S

Sample Output 1:

nonoyesThere are 2 components.

Sample Input 2:

5C 3 2I 3 2C 1 5I 4 5I 2 4C 3 5I 1 3C 1 5S

Sample Output 2:

nonoyesyes
The network is connected.
思路：本题的关键点是按秩归并和路径压缩，关于这两点在MOOC上老师有很详细的解释。
#include<stdio.h>#include<stdlib.h>#include<algorithm>using namespace std;int S[10001];int N;int Find(int S[],int i){if(S[i] < 0){return i;}elsereturn  S[i] = Find(S,S[i]);/*不断的找父结点，一直找到根结点，然后再把每个父结点指向根结点*/}void check(int S[]){int root1,root2;int a, b;scanf("%d %d",&a,&b);root1 = Find(S,a-1);root2 = Find(S,b-1);if(root1 == root2){printf("yes\n");}elseprintf("no\n");}void Union(int S[],int root1,int root2){if(S[root2] < S[root1]){S[root2] += S[root1];S[root1] = root2;}else{S[root1] += S[root2];S[root2] = root1;}}void Insert(int S[]){int root1,root2;int a, b;scanf("%d %d",&a,&b);root1 = Find(S,a-1);root2 = Find(S,b-1);if(root1 != root2){Union(S,root1,root2);}}int main(void){char ch;int count = 0;scanf("%d",&N);for(int i = 0; i < N; i++)/*初始化令每个结点都是根结点*/{S[i] = -1;}while(1){scanf("%s",&ch);if(ch == 'C'){check(S);}else if(ch == 'I'){Insert(S);}else if(ch == 'S'){break;}}for(int i = 0; i < N; i++){if(S[i] < 0)count++;}if(count == 1){printf("The network is connected.");}else{printf("There are %d components.",count);}return 0;}




在之前的Find函数中我用的是循环来返回根结点，结果犯了一个错误。
int Find(int S[],int i)
{
for(;S[i] > 0; i = S[i]);
return i;
}
for循环后的那个分号我没加，结果导致i = S[i]这条语句在for循环判断后（不满足S[i] > 0）,仍然被执行。