POJ - 1611 The Suspects 简单并查集

本文介绍了一个基于并查集算法的程序设计案例,该程序用于模拟SARS在大学校园内潜在的传播路径,并确定可能受感染的学生数量。

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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

大致题意:有n个学生,有m组,学生编号从0到n-1。已知编号为0的学生可能患有SARS,与可能患有SARS的学生同一组的其他学生都有可能患有SARS,问n个学生里有多少学生可能患有SARS。

思路:简单的并查集。

代码如下

#include<iostream>
#include<algorithm>
#include<string>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<cstring>
using namespace std;
const int maxn=30005;
int pre[maxn];

int find(int x)
{
   int r=x;
   while (pre[r]!=r)
   r=pre[r];
   int i=x; int j;
   while(i!=r)
   {
       j=pre[i];
       pre[i]=r;
       i=j;
   }
   return r;
}

void join(int x,int y)
{
    int f1=find(x);
    int f2=find(y);
    if(f1!=f2)
    {
        pre[f2]=f1;
        //total--;
    }
}

int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0&&m==0) break;

        for(int i=0;i<=n;i++)
        pre[i]=i;
        while(m--)
        {
            int k;
            scanf("%d",&k);
            int x,y;
            scanf("%d",&x);
            for(int i=1;i<k;i++)
            {
                scanf("%d",&y);
                join(x,y);
            }
        }
        int sum=1;
        for(int i=1;i<=n;i++)
        {
            if(find(0)==find(i))
            sum++;
        }
        printf("%d\n",sum);
    }
   return 0;
}
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