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The Suspects
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 Sample Output 4 Source |
题目大意 ：有n个学生 m个小组，每小组有k个学生，一个学生可能会參加多个小组。小组中仅仅要有一个人得病，其余的都是嫌疑人。如今已知这些小组的人员，且0号学生已经患病。求一共同拥有多少个嫌疑人。
并查集的基本运用
每一个小组的人员属于同一个集合。在根节点记录每一个集合的个数num[i]，然后找到0所属的根节点。num[find_Set(0)]即为全部的嫌疑人数。
#include <iostream>
#include<cstdio>
using namespace std;
const int Max=300010; int p[Max],rank[Max];
int num[Max];//元素个数 int Find_Set(int x)
{
if (x!=p[x])
p[x]=Find_Set(p[x]);
return p[x];
} void Union(int x,int y)
{
x=Find_Set(x);
y=Find_Set(y);
if(x==y)
return;
if (rank[x]>rank[y])
{
p[y]=x;
num[x]+=num[y];
}
else
{
p[x]=y;
if (rank[x]==rank[y])
rank[x]++;
num[y]+=num[x];
}
} int main()
{
int i,j,k,n,m;
int a,b;
int s1,s2;
while (cin>>n>>m)
{
if(n==0&&m==0)
{
break;
}
for (i=0;i<n;i++)
{
p[i]=i;
rank[i]=0;
num[i]=1;
}
for (i=0;i<m;i++)
{
cin>>k>>a;
for (j=1;j<k;j++)
{
cin>>b;
s1=Find_Set(a);
s2=Find_Set(b);
Union(s1,s2);
}
}
i=Find_Set(0);
cout<<num[i]<<endl;
}
return 0;
}