The Suspects 并查集
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
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
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
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <string>
#include <cmath>
using namespace std;
int n,m,f[],k,ex[],c;
int init()
{
for(int i=;i<n;i++)
f[i]=i;
}
int getf(int x)
{
if(f[x]!=x)f[x]=getf(f[x]);
return f[x];
}
int merge(int x,int y)
{
int xx=getf(x),yy=getf(y);
f[yy]=xx;
}
int main()
{
while(scanf("%d%d",&n,&m))
{
c=;
if(!m&&!n)break;
init();
for(int i=;i<m;i++)
{
scanf("%d",&k);
for(int i=;i<k;i++)
{
scanf("%d",&ex[i]);
if(i>)merge(ex[i-],ex[i]);
}
}
int fa=getf();
for(int i=;i<n;i++)
if(getf(i)==fa)c++;
cout<<c<<endl;
}
}
The Suspects 并查集的更多相关文章
- The Suspects(并查集维护根节点信息)
The Suspects Time Limit: 1000MS Memory Limit: 20000K Total Submissions: 37090 Accepted: 17980 De ...
- poj 1611 The Suspects(并查集输出集合个数)
Description Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, wa ...
- poj 1611 The Suspects 并查集变形题目
The Suspects Time Limit: 1000MS Memory Limit: 20000K Total Submissions: 20596 Accepted: 9998 D ...
- B - The Suspects(并查集)
B - The Suspects Time Limit:1000MS Memory Limit:20000KB 64bit IO Format:%lld & %llu Desc ...
- POJ 1611 The Suspects (并查集+数组记录子孙个数 )
The Suspects Time Limit: 1000MS Memory Limit: 20000K Total Submissions: 24134 Accepted: 11787 De ...
- POJ 1611 The Suspects (并查集求数量)
Description Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, wa ...
- POJ 1611 The Suspects 并查集 Union Find
本题也是个标准的并查集题解. 操作完并查集之后,就是要找和0节点在同一个集合的元素有多少. 注意这个操作,须要先找到0的父母节点.然后查找有多少个节点的额父母节点和0的父母节点同样. 这个时候须要对每 ...
- poj 1611 The Suspects 并查集
The Suspects Time Limit: 1000MS Memory Limit: 20000K Total Submissions: 30522 Accepted: 14836 De ...
- The Suspects(并查集求节点数)
The Suspects Time Limit: 1000MS Memory Limit: 20000K Total Submissions: 28164 Accepted: 13718 De ...
- [ACM] POJ 1611 The Suspects (并查集,输出第i个人所在集合的总人数)
The Suspects Time Limit: 1000MS Memory Limit: 20000K Total Submissions: 21586 Accepted: 10456 De ...
随机推荐
- cygwin install git
Installation with Cygwin If you're comfortable with Cygwin, then use it to install git, ssh, wget an ...
- Python mysql-表中数据的大量插入
2017-09-06 23:28:26 import pymysql db = pymysql.connect("localhost","root"," ...
- English trip -- 国际音标表
26个字母音标表 A a [ei] B b [bi:] C c [si:] D d [di:] E e [i:] F f [ef] G g [dʒi:] H h [eit∫] I i [ai] J j ...
- 源代码方式调试Mycat
如果是第一次刚接触MyCat建议下载源码在本地通过eclipse等工具进行配置和运行,便于深入了解和调试程序运行逻辑. 1)源代码方式调试与配置 由于MyCat源代码目前主要托管在github上,大家 ...
- LeetCode--171--Excel表列序号
问题描述: 给定一个Excel表格中的列名称,返回其相应的列序号. 例如, A -> 1 B -> 2 C -> 3 ... Z -> 26 AA -> 27 AB -& ...
- Rspec: feature spec 功能测试 测试JavaScript.
我们要把应用各组件放在一起做集成 测试,这样才能保证模型和控制器之间能够良好契合. 在 RSpec 中,这种测试称为功能测试(feature spec),有时也称为验收测试(acceptance te ...
- Knapsack CodeForces - 1132E (多重背包)
可以将大量同种物品合并为$lcm$来优化, 复杂度$O(nlcm^2)$, 好像可以用bitset优化到$O(nlcm^2/\omega)$, 但是没看太懂 const int L = 840, M ...
- Python在七牛云平台的应用(二)图片瘦身
(一)七牛云平台的图片瘦身功能简介:(引用自官网) 针对jpeg.png格式图片 瘦身后分辨率不变,格式不变. 肉眼画质不变. 图片体积大幅减少,节省 CDN 流量 官网给的图片压缩率很高,官网给的「 ...
- codefroces 450B矩阵快速幂
找出递推关系式就好了 (fi+1)=(1 -1)(fi ) ( fi)=(1 0)(fi-1) 不会打矩阵将就着看吧... 这是第一道矩阵快速幂.细节还是有很多没注意到的 本来想看挑战写 ...
- exec可以用来执行语句的
set @sql='select * from '+@table print @sql exec(@sql)