The Suspects||POJ1611-CSDN博客

本文链接：https://blog.csdn.net/csdn_muxin/article/details/76595555

本文提供了一道名为“The Suspects”的POJ 1611题目解答，采用并查集算法来解决学生群体中疑似病例的追踪问题。通过输入学生群体信息，程序能够快速识别所有潜在的接触者。

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题目链接：http://poj.org/problem?id=1611
The Suspects

Time Limit: 1000MS      Memory Limit: 20000K
Total Submissions: 40525        Accepted: 19575

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

题解：简单的并查集运用

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int par[30003],num[30003];
void initial(int n)//初始化，每个人的父节点是自己，每棵树大小都为1 
{
    int i;
    for(i=0;i<n;i++)
    {
        par[i]=i;
        num[i]=1;
    }
}
int find(int x)//找祖宗 
{
    if(x!=par[x])
      par[x]=find(par[x]);
    return par[x];
}
void unite(int x,int y)//合并 
{
    int fx=find(x);
    int fy=find(y);
    if(fx==fy)//祖宗相同跳出 
      return ;
    if(num[fx]>=num[fy])//孩子多的是爸爸 
    {
        par[fy]=fx;
        num[fx]+=num[fy];
    }
    else
    {
        par[fx]=fy;
        num[fy]+=num[fx];
    }
}
int main()
{
    int n,m,k,a,b,i,j;
    while(cin>>n>>m)
    {
        if(n==0&&m==0) break;
        initial(n);
        for(i=0;i<m;i++)
        {
            scanf("%d%d",&k,&a);
            for(j=1;j<k;j++)
            {
                scanf("%d",&b);
                unite(a,b);
            }
        }
        printf("%d\n",num[par[0]]);//par【0】，是0的父节点，可能是0本身，然后num【par【0】】其实就是0所在这棵树的大小 
    }
return 0;
}