The Suspects POJ 1611

最新推荐文章于 2021-08-09 06:56:03 发布

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原文链接：http://www.cnblogs.com/-xiangyang/p/9220263.html

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#数据结构与算法

本文介绍了一种通过学生群体间关系模拟SARS潜在传播范围的算法。在一个假设的大学环境中，通过建立学生群体间的联系来确定可能的最大感染人数。利用并查集数据结构实现了高效的成员查找和群组合并。

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The Suspects

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

题意： 0号必然感染，和0号一组的是感染者，还有这一组中的去其他人也会去感染其他人，然后问你最多有多少感染者

题解：有一个技巧就是初始化是t[i]=1；在mix（）函数中相加。需要注意此题最少有一个人。

#include<stdio.h>
int road[10000000+10];
int t[10000000+10]; 
int find(int a)
{
    if(road[a]==a) return a;
    else
    return road[a]=find(road[a]);
}
void mix(int a,int b)
{

    int x;
    int y;
    x=find(a);
    y=find(b);
    if(x!=y)
    {
        road[y]=x;
        t[x]+=t[y];//是两个集合连接在一起，并使人数相加
    }

}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    if(n==0&&m==0)
    {
        break;
    }
    else
    {
        int a, b,c,maxx=0;
        for(int i=0;i<=n;i++)
        {
            road[i]=i;
            t[i]=1;
        }
        while(m--)
        {

            scanf("%d",&a);
            scanf("%d",&b);
            for(int i=0;i<a-1;i++)
            {
                scanf("%d",&c);
                mix(c,b);
            }
        }
        int MAX=0;  
        MAX=t[find(0)];
        printf("%d\n",MAX);
    }
    return 0;
}