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POJ 1611 The Suspects

2024-07-12 06:58:35   219人阅读

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
题意：有最多30000的学生被怀疑得了SARS病，而他们又在不同的社团里面，只要这个社团有1人被怀疑得了病，那么这个社团全部的人将会被怀疑得了病。现在给你总共学生的数量  跟社团数  接下来输入每个社团成员的学号，每行输入的第一个数为该社团人的数量。要你求最多可能有多少人得了病
思路：将每个社团的人放在同一个节点，接下来遍历1到N（可以优化），统计下所有可能得病的人即可
AC代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 30005
int f[N],sum;
int find(int x)
{
    if(x!=f[x])
        f[x]=find(f[x]);
    return f[x];
}
void make(int x,int y)
{
    int a,b;
    a=find(x);
    b=find(y);
    if(a!=b)
        f[a]=b;
}
int main()
{
    int n,m,i;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        if(n==0&&m==0)break;
        sum=1;
        for(i=0;i<N;i++)
            f[i]=i;
        while(m--)
        {
            int k,a,b,t;
            scanf("%d",&k);
            scanf("%d",&t);
            for(i=1;i<k;i++){
                 scanf("%d",&b);
                    make(b,t);
            }
        }
        for(i=0;i<N;i++){
            if(find(0)==find(i))
            {
                if(i!=find(i))
                    sum++;
            }
        }
        printf("%d\n",sum);
    }

    return 0;
}



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