数据结构 05-树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≤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
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.
并查集的查找与合并
参考文章 https://zhuanlan.zhihu.com/p/93647900/
#include <iostream> #include <vector> using namespace std; class unionFindSet{ public: vector<int> ufs; vector<int> rank; unionFindSet()=default; unionFindSet(int n){ ufs=vector<int>(n+1); rank=vector<int>(n+1,1); for(int i=0;i<n+1;i++){ ufs[i]=i; } }; int find(int x){ return ufs[x]==x?x:ufs[x]=find(ufs[x]); } void merge(int i,int j){ int x=find(i),y=find(j); if(rank[x]<=rank[y]){ ufs[x]=y; }else{ ufs[y]=x; } if(rank[x]==rank[y]&&x!=y){ rank[y]++; } } void components(){ int count{0}; for(int i=1;i<ufs.size();i++){ if(ufs[i]==i){ count++; } } if(count==1){ cout <<"The network is connected."<<endl; }else{ cout <<"There are "<< count <<" components."<<endl; } } void judge(){ int x,y; string str; cin >> str; while(str!="S"){ cin >> x >> y; if(str=="I"){ merge(find(x), find(y)); } if(str=="C"){ if(find(x)==find(y)){ cout << "yes"<<endl; }else{ cout << "no"<<endl; } } cin >> str; } } }; int main(){ int n; cin >> n; unionFindSet ufs{n}; ufs.judge(); ufs.components(); return 0; }