【原创】poj ----- 1611 The Suspects 解题报告

题目地址：

http://poj.org/problem?id=1611

 

题目内容：

The Suspects

Time Limit: 1000MS		Memory Limit: 20000K
Total Submissions: 24253		Accepted: 11868

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

Source

Asia Kaohsiung 2003

 

解题思路：

 

稍微有点变化的并查集，但还算水题。

 

关键在于：

0、0号同学肯定有病。

1、必须记录集合的大小已供输出。

 

因此，我们合并每组的同学，最后找到0所在集合的大小并输出结果。

 

维护集合大小的时候，可以只维护根节点的子节点数目。举个例子，我们用数组num来保存每个集合的大小，假如2，3，4，5四个人位于同一个集合，而且2是根节点，那么只有num[2]的值是有意义的，说明了2拥有的子节点数目。其余的例如num[3]、num[4]等值都没有意义。

而这个集合的大小便是num[2] + 1

 

全部代码：

#include <stdio.h>

int man[30001];
int num[30001]; // 记录有几个子节点，但只有根记录的数据才算数

void init(int n)
{
  for (int i = 0; i < n; i ++) {
    man[i] = -1;
    num[i] = 0;
  }
}

int find_root(int child)
{
  if (man[child] == -1)
    return child;
  return man[child] = find_root(man[child]);
}

void union_set(int one, int two) 
{
  int fat1 = find_root(one);
  int fat2 = find_root(two);
  //printf("%d and %d has been union. ", fat1, fat2);
  if (fat1 == fat2)
    return;
  num[fat1] += num[fat2] + 1;
  man[fat2] = fat1;
  //printf("father is %d\n", fat1);
}

int main(void)
{
  int m,n;
  while (scanf("%d%d", &n, &m)) {
    if (m == 0 && n == 0)
      break;
    init(n);
    for (int i = 0; i < m; i ++) {
      int count,pre,now;
      scanf("%d", &count);
      pre = -1;
      for (int j = 0; j < count; j ++) {
    scanf("%d", &now);
    if (pre != -1) {
      union_set(pre, now);
    }
    pre = now;
      }
    }
    int fa = find_root(0);
    
    printf("%d\n", num[fa] + 1);
  }
  return 0;
}