PTA 中國大學MOOC-陳越、何欽銘-資料結構-2018秋 05-樹8 File Transfer (25 分) 並查集
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
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.
程式碼如下:
/*
並查集的查詢與合併
一開始初始化集合的個數為n個..
然後每兩個集合合併就將集合數減一..
最後判斷集合個數是否為一來判定是否兩兩相互連線...
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=10005;
int n;
int a[maxn];
int num;
char op[5];
void init()
{
num=n;
for (int i=1;i<=n;i++)
{
a[i]=i;
}
}
int Find(int x)
{
if(a[x]==x)
return x;
else
return a[x]=Find(a[x]);
}
void unit(int x,int y)
{
int px=Find(x);
int py=Find(y);
if(px!=py)
{
a[px]=py;
num--;
}
}
int main()
{
scanf("%d",&n);
init();
while (scanf("%s",op)&&op[0]!='S')
{
int x,y;
scanf("%d%d",&x,&y);
if(op[0]=='I')
{
unit(x,y);
}
else
{
if(Find(x)==Find(y))
{
printf("yes\n");
}
else
printf("no\n");
}
}
if(num==1)
printf("The network is connected.\n");
else
printf("There are %d components.\n",num);
return 0;
}