POJ 1611 The Suspects 并查集

N - The Suspects

Time Limit:1000MS     Memory Limit:20000KB     64bit IO Format:%lld & %llu

Submit Status

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 42 1 25 10 13 11 12 142 0 12 99 2200 21 55 1 2 3 4 51 00 0

Sample Output

411

题意:需找SARS病毒感染者嫌疑人数，凡与感染者0号同学有同组关系的皆为嫌疑人。

思路:凡是有分堆的问题一般都会想到并查集,题目中说只要和0同学有关皆为嫌疑人,那么我们在进行并查集合并的时候就需要判断一下要合并的

两个同学之间有没有0号同学,如果有的话就让他指向0号同学如果没有则直接合并,最后统计祖先为0号同学的人数即可;

#include<stdio.h>#include<string.h>int pre[30003];int find(int x){  int r=x;  while(pre[r]!=r)    r=pre[r];    int i=x,j;    while(pre[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)   { if(f1!=0)     pre[f1]=f2;     else     pre[f2]=f1;   }   return ;}int main(){   int n,m;    int k,a[30003];    int sum;    while(scanf("%d %d",&n,&m)!=EOF)    { if(m==0&&n==0)       break;       sum=0;      for(int i=0;i<n;i++)       pre[i]=i;      while(m--)      { scanf("%d",&k);        for(int i=0;i<k;i++)          {           scanf("%d",&a[i]);           join(a[0],a[i]);}//因为一组中有多个同学,直接让第一个和后面所有的合并;                        }       for(int i=0;i<n;i++)        if(find(i)==0)        sum++;        printf("%d\n",sum);    }    return 0;}