POJ 【1611】The Suspects

最新推荐文章于 2019-10-06 08:12:25 发布

weixin_34387284

最新推荐文章于 2019-10-06 08:12:25 发布

阅读量53

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/jiangyongy/p/3971677.html

这篇文章介绍了一个算法，用于在一个大学中识别所有可能感染非传染性疾病的学生群体的成员。通过输入学生总数、学生群体数量以及每个群体的成员列表，该算法能够找出所有疑似病例，并输出其数量。

The Suspects
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

这道题的大意为：先输入n，m，接下来的m行，每行第一个为这一行的数个数，每一行为一个集合，最后输出以0为根节点的集合有几个。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int fa[40010];
int num[40010];
int fin(int a)//查找根节点的函数
{
    if(a!=fa[a])
        fa[a]=fin(fa[a]);//路径压缩
    return fa[a];
}
void un(int a,int b)//集合合并的函数
{
    int x=fin(a);
    int y=fin(b);
    if(x!=y)
        fa[y]=x;
}
int main()
{
    int n,n1,m,i,j,a,b;
    while(scanf("%d %d",&n,&m))
    {
        if(n==0)
            return 0;
        memset(num,0,sizeof(num));
        int flag=0;
        for(i=0; i<n; i++)
            fa[i]=i;
        for(i=0; i<m; i++)
        {
            scanf("%d",&n1);
            scanf("%d",&a);
            for(j=1; j<n1; j++)//让每行第一个数后的数都与第一个数进行合并
            {
                scanf("%d",&b);
                un(a,b);
            }
        }
       for(i=0;i<n;i++)
       {
           if(fin(i)==fin(0))//找的以0为根节点的集合
            flag++;
       }
        printf("%d\n",flag);
    }
    return 0;
}