poj 1611 The Suspects 并查集变形题目

The Suspects

 

Time Limit: 1000MS		Memory Limit: 20000K
Total Submissions: 20596		Accepted: 9998

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
讲解：使用并查集，每输入一行学生编号数据后，都判断第一个学生编号的祖先是否和后面每一个学生编号的祖先相同，若不同，则将第一个编号的祖先变为该后面编号的祖先，
     将集合合并，且每合并一次，第一个编号的祖先上的元素个数都等于合并之前两个祖先上的元素个数之和（最初，每一个编号的祖先都是该编号自身，且对应的元素个数都为一），
     最后输出编号为0的祖先上的元素个数即可 
总结及出错情况：该题主要就是使用并查集，注意在合并的时候祖先上的元素个数要更新，且最初的时候每一个编号的祖先上的元素个数都是1，最易出错的就是在输入的数据中没有0
     就以为没有感染嫌疑者，其实这种情况应该是有一个，因为0在的号代表的学生就是感染嫌疑者
AC代码：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define N 30010
int vis[N],parent[N],m,n;
void begin()
{
    for(int i=0;i<m;i++)
    {
         parent[i]=i;
         vis[i]=1;
    }
}
int find (int x)
{
    if(parent[x]!=x)
    {
        parent[x]=find(parent[x]);
    }
    return parent[x];
}
int query(int x,int y)
{
    int px=find(x);
    int py=find(y);
    if(px!=py)
    {
        parent[py]=px;
        vis[px]=vis[py]+vis[px];//父亲不同的话统计个数，相同的话，不用统计了；
    }
}
int main()
{
    int first,k,a;
    while(cin>>m>>n && m+n)
    {
        begin();
        for(int i=0; i<n; i++)
        {
            scanf("%d %d", &k,&first);         //  首先把第一个输入的，当成父亲；
            for(int j=1; j<k; j++)
            {
                scanf("%d",&a);               //后面的如果有父亲，并且不和第一个父亲相同，
                   query(first ,a);           //则后面的父亲，为第一个的父亲
            }
        }
    printf("%d
",vis[find(0)]);
    }
    return 0;
}