顾名思义,并查集就是“并”和”查“,实际上,并查集通常开始要初始化为自己本身,然后再查找元素所在的集合,即根节点,再将两个元素合并成一个元素(合并之前应该判断两个元素是否同属一个集合,可以用查找函数实现)。
例题: 

Description

Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. To minimize transmission to others, the best strategy is to separate the suspects from others. 
In the Not-Spreading-Your-Sickness University (NSYSU), there are many student groups. Students in the same group intercommunicate with each other frequently, and a student may join several groups. To prevent the possible transmissions of SARS, the NSYSU collects the member lists of all student groups, and makes the following rule in their standard operation procedure (SOP). 
Once a member in a group is a suspect, all members in the group are suspects. 
However, they find that it is not easy to identify all the suspects when a student is recognized as a suspect. Your job is to write a program which finds all the suspects.

Input

The input file contains several cases. Each test case begins with two integers n and m in a line, where n is the number of students, and m is the number of groups. You may assume that 0 < n <= 30000 and 0 <= m <= 500. Every student is numbered by a unique integer between 0 and n−1, and initially student 0 is recognized as a suspect in all the cases. This line is followed by m member lists of the groups, one line per group. Each line begins with an integer k by itself representing the number of members in the group. Following the number of members, there are k integers representing the students in this group. All the integers in a line are separated by at least one space. 
A case with n = 0 and m = 0 indicates the end of the input, and need not be processed.

Output

For each case, output the number of suspects in one line.

Sample Input

100 4

2 1 2
5 10 13 11 12 14
2 0 1
2 99 2
200 2
1 5
5 1 2 3 4 5
1 0
0 0

 Sample Output
4
1
1

这个题目的大意就是从0号开始往后寻找(默认0号是被传染的),找到和0号相关的人,看看总共有多少人与0
有直接或间接有联系,然后输出值,题意可能不太好理解,题目还是很简单的,可以用并查集做。

#include<stdio.h>
int bin[30001],rank[30001];
int find(int x)
{//查找函数
 int r=x;
 while(bin[r]!=r)
  r=bin[r];
 return r;
}
void find1(int x,int y)
{//合并函数
 int fx,fy;
 fx=find(x);
 fy=find(y);
 if(fx==fy)
  return;
 if(rank[fx]>rank[fy])
 {
  bin[fy]=fx;
  rank[fx]+=rank[fy];
 }
 else
 {
  bin[fx]=fy;
  rank[fy]+=rank[fx];
 }
}
int main()
{
 int m,n;
 while(scanf("%d%d",&n,&m)!=EOF)
 {
  int i,count=0,k,j,d,a;
  if(m==0&&n==0)
   break;
  for(i=0;i<=n;i++)
  {
   bin[i]=i;
   rank[i]=1;
  }
  for(i=0;i<m;i++)
  {
   scanf("%d%d",&k,&d);
   for(j=1;j<k;j++)
   {
    scanf("%d",&a);
     find1(d,a);
   }
  }
  printf("%d\n",rank[find(0)]);
 }
 return 0;
}


 

 

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