poj 1611 The Suspects

最新推荐文章于 2019-10-06 08:12:25 发布

w__gavin

最新推荐文章于 2019-10-06 08:12:25 发布

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分类专栏：并查集

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本文介绍了一个基于并查集算法的学生群体SARS传播模拟程序。该程序通过输入学生群体信息及初始疑似病例，找出所有潜在的感染风险人群。适用于研究传染疾病在特定社会网络中的扩散规律。

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The Suspects
Time Limit: 1000MS Memory Limit: 20000K
Total Submissions: 42609 Accepted: 20555
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

基础并查集

#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <cmath>
using namespace std;
int pre[35000];
int findroot(int x)
{
    int r=x;
    while(pre[r]!=r)
    {
        r=pre[r];
    }
    int i = x;
    while(i != r)
    {
        int j;
        j =  pre[i];
        pre[i] = r;
        i = j;
    }
    return r;
}
void mix(int a,int b)
{
    int roota = findroot(a);
    int rootb = findroot(b);
    if(roota != rootb)
    {
        pre[roota] = rootb;
    }
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m) && m+n != 0)
    {
        for(int i = 0; i < n; i++)
        {
            pre[i] = i;
        }
        for(int i = 0; i < m; i++)
        {
            int nn;
            int temp;
            scanf("%d",&nn);
            scanf("%d",&temp);
            for(int i = 0; i < nn-1; i++)
            {
                int a;
                scanf("%d",&a);
                mix(temp,a);
            }
        }
        int ans = findroot(0);
        int sum = 0;
        for(int i = 0; i < n; i++)
        {
            if(findroot(i) == ans)
            {
                sum++;
            }
        }
        printf("%d\n",sum);
    }
}