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最新推荐文章于 2022-07-13 14:39:59 发布

AOATH.

最新推荐文章于 2022-07-13 14:39:59 发布

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本文链接：https://blog.youkuaiyun.com/weixin_43976373/article/details/86686902

本文解析了一道经典的并查集算法题目POJ1611，介绍了如何通过并查集算法解决学生群体中疑似病例的识别问题。文章详细解释了输入输出格式，给出了完整的C++代码实现，并说明了并查集的基本操作，如初始化、查找和合并。

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POJ 1611

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

最基础的并查集题了
可以用一个数组来记录根节点的总人数



#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>

using namespace std;

 int maxn = 30000;

int  per[maxn], num[maxn], n, m;

int a[maxn];

void
init(){

    for(int i = 0; i < n ; ++i){

        per[i] = i;

        num[i] = 1;

    }

    return ;

}

int find(int x){

    if(x == per[x])

        return x;

    return per[x] = find(per[x]);

}

void join(int x, int y){

    int fx = find(x);

    int fy = find(y);

    if(fx != fy){

        per[fx] = fy;

        num[fy] += num[fx];

    }

    return ;

}

int
main (){

    while(scanf("%d%d", &n,&m), n || m){

            init();

        while(m--){

            int t;

            scanf("%d", &t);

            for(int i = 0; i < t; ++i){

                scanf("%d",&a[i]);

            }

            for(int i = 0 ; i < t - 1; ++i)

                join(a[i], a[i + 1]);

        }

        int  k = find(0);

        printf("%d\n", num[k]);

    }

    return 0;

}