File Transfer(25) 与朋友圈(25)（简单并查集）

7-1 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 $N$ ( $2 \leq N \leq 1 0^{ 4 }$ ), 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

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.

题目大意：输入为c时查询两个数是否相同，为s时输出最大联通数。

思路：裸并查集

代码如下：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<math.h>
using namespace std;
int p[1000000];
void int1(int n)
{
 for(int i=1;i<=2*n;i++)
 p[i]=i;
} 
int find(int x)
{
 int y=x,t;
 while(y!=p[y]) y=p[y];
 while(x!=y)
 {
  t=p[x];p[x]=y;x=t;
 }
 return x;
}
void bin(int x,int y)
{
 int a=find(x);
 int b=find(y);
 if(a!=b)
 {
  if(p[a]<p[b])
  {
   p[b]=a;
  }else
  {
   p[a]=b;
  }
 }
}
int main()
{
     int t,i,j,a,b;
     char ch;
     scanf("%d",&t);
     int1(t);
     while(scanf("%c",&ch)!=EOF&&ch!='S')
     {
      if(ch=='I')
      {
       scanf("%d%d",&a,&b);
       bin(a,b);//注意此处若改为p[a]=b会错，原因是未进行路径压缩
  }
  if(ch=='C')
  {
   scanf("%d%d",&a,&b);
   if(find(a)==find(b))
   printf("yes\n");
   else printf("no\n");
  }
 }
 int s=0;
 for(i=1;i<=t;i++)
 {
  if(p[i]==i)
  s++;
 }
 if(s==1) printf("The network is connected.\n");
 else printf("There are %d components.\n",s);
 return 0;
}

7-2 朋友圈（25 分）

某学校有N个学生，形成M个俱乐部。每个俱乐部里的学生有着一定相似的兴趣爱好，形成一个朋友圈。一个学生可以同时属于若干个不同的俱乐部。根据“我的朋友的朋友也是我的朋友”这个推论可以得出，如果A和B是朋友，且B和C是朋友，则A和C也是朋友。请编写程序计算最大朋友圈中有多少人。

输入格式:

输入的第一行包含两个正整数N（ $\leq$ 30000）和M（ $\leq$ 1000），分别代表学校的学生总数和俱乐部的个数。后面的M行每行按以下格式给出1个俱乐部的信息，其中学生从1~N编号：

第i个俱乐部的人数Mi（空格）学生1（空格）学生2 … 学生Mi

输出格式:

输出给出一个整数，表示在最大朋友圈中有多少人。

输入样例:

输出样例:

思路：还是裸的并查集

代码如下：

#include<iostream>
#include<string.h>
using namespace std;
int par[30001],num[30001];
int fi(int x)
{
    int r=x;
    while(r!=par[r])
        r=par[r];
    int i=x,j;
    while(i!=r)
    {
        j=par[i];
        par[i]=r;
        i=j;
    }
    return r;
}
void Join(int x,int y)
{
    int fx=fi(x);
    int fy=fi(y);
    if(fx!=fy)
        par[fx]=fy;
}
int  main()
{
    std::ios::sync_with_stdio(false);
    int n,m,a,b,k;
    cin>>n>>m;
    for(int i=1;i<=n;i++)
        par[i]=i;
    while(m--)
    {
        cin>>k;
        cin>>a;
        k--;
        while(k--)
        {
            cin>>b;
            Join(a,b);
        }
    }
    memset(num,0,sizeof(num));
    for(int i=1;i<=n;i++)
    {
        num[fi(i)]++;
    }
    int ans=0;
    for(int i=1;i<=n;i++)
    {
        if(num[i]>ans)
            ans=num[i];
    }
    cout<<ans<<endl;
    return 0;
}