POJ 1611 The Suspects - 并查集

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
这道题就是说0号童鞋是有患病嫌疑的，和0号童鞋一个组的也有患病嫌疑，和有患病嫌疑在一个组的同学也有患病嫌疑，问一共有几个孩子有患病嫌疑。
首先，可以把一个组（更准确的说是被传染的）的同学放在一个集合里，然后记录下这个集合的人数，最后找到0童鞋所在的集合的人数。

#include <iostream>
#include<stdio.h>
# define max 30005
using namespace std;

int rank[max]; //秩
int num[max];//记录一个集合有多少人
int fa[max];//记录父亲结点

int find(int x){
 int father=fa[x];
  if(x!=fa[x])
  {fa[x]=find(fa[father]);
  return fa[x];}
  else return father;
}

void link(int a,int b){
  int x=find(a);
  int y=find(b);
  if(x==y) return;//如果根结点一致，说明之前已经添加过，不必重复添加
  if(rank[x]>rank[y])//x的秩比y的大，把y连接到x上
   {fa[y]=x;
   num[x]+=num[y];}//连接后记得修改总人数
   else
   {if(rank[x]==rank[y])
    rank[y]++;
   fa[x]=y;
   num[y]+=num[x];}
}

int main()
{
   int n,m,i;
   while(scanf("%d %d",&n,&m)!=EOF)
   {
       if(n+m==0)break;
     for(i=0;i<n;i++){
        rank[i]=1;
        num[i]=1;
        fa[i]=i;
    }
    int k,x,y;
    for(i=0;i<m;i++){
        scanf("%d",&k);
        scanf("%d",&x);
      for(int j=1;j<k;j++){
         scanf("%d",&y);
         link(x,y); //相邻两个进行连接
         x=y;
     }
    }
    int root=find(0);//找到0童鞋所在的根结点
    int ans=num[root];
    if(m==0)printf("1\n");
    else printf("%d\n",ans);
   }
       return 0;
}