1611 The Suspects

The Suspects
Time Limit: 1000MS Memory Limit: 20000K
Total Submissions: 7763 Accepted: 3683

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
#include<iostream>
using namespace std;
int father[50002],num[50002];
int find(int x)
{
    if(father[x]!=x)
    {
        father[x]=find(father[x]);
    }
    return father[x];
}
void uion(int x,int y)
{
    x=find(x);
    y=find(y);
    if(x==y) return ;
    if(num[x]>num[y])
    {
        father[y]=x;
        num[x]+=num[y];
    }
    else
    {
        father[x]=y;
        num[y]+=num[x];
     }       
}           
int main()
{
    int n,m;
    while(cin>>n>>m,n||m)
    {
        for(int i=1;i<=n;i++) father[i]=i,num[i]=1;
        while(m--)
        {
            int p;
            cin>>p;
            if(p==0) continue;
            int x,y;
            cin>>x;x++;p--;
            while(p--)
            {
                cin>>y;y++;
                uion(x,y);
            }   
        }
        int countc=find(1);
        cout<<num[countc]<<endl;
    }
    return 0;
}           
任务描述 本关任务:求最喜欢看电影(影评次数最多)的那位女性评最高分的 10 部电影(观影者,电影名,评分)。 相关知识(略) 编程要求 根据提示,在右侧编辑器补充代码,求最喜欢看电影(影评次数最多)的那位女性评最高分的 10 部电影,若评分相同,则按电影名称降序排序(观影者,电影名,评分)。 使用数据库:mydb 用户表:t_user 字段名 类型 注释 userid bigint 用户ID sex string 性别 age int 年龄 occupation string 职业 zipcode string 邮政编码 部分数据如下: 1 F 1 10 48067 2 M 56 16 70072 3 M 25 15 55117 用户表:t_movies 字段名 类型 注释 movieid bigint 电影ID moviename string 电影名字 movietype string 电影类型 部分数据如下: 1 Toy Story (1995) Animation|Children's|Comedy 2 Jumanji (1995) Adventure|Children's|Fantasy 3 Grumpier Old Men (1995) Comedy|Romance 用户表:t_ratings 字段名 类型 注释 userid bigint 用户ID movieid bigint 电影ID rate double 评分 times string 评分时间戳 部分数据如下: 1 48 5.0 978824351 1 150 5.0 978301777 1 1 5.0 978824268 测试说明 平台会对你编写的代码进行测试: 预期输出: 151 Usual Suspects, The (1995) 5.0 151 Twelve Monkeys (1995) 5.0 151 Things to Do in Denver when You're Dead (1995) 5.0 151 Strange Days (1995) 5.0 151 Persuasion (1995) 5.0 151 Leaving Las Vegas (1995) 5.0 151 Braveheart (1995) 5.0 151 To Die For (1995) 4.0 151 Sense and Sensibility (1995) 4.0 151 Rob Roy (1995) 4.0
最新发布
06-18
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