并查集poj611suspect
来源:互联网 发布:人工智能的好处 编辑:程序博客网 时间:2024/06/03 20:33
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
#include<iostream>using namespace std;int father[30000],rank[30000];int n,m;void init(){ for(int i=0;i<n;i++) { father[i]=i; rank[i]=1; }}int find(int x){ if(x!=father[x]) { father[x]=find(father[x]);//之所以不可以直接返回是因为这里可能有多层的关系,优化 } return father[x];}void uniont(int x,int y){ int xx=find(x); int yy=find(y); if(xx==yy) return ; else { if(rank[xx]>=rank[yy]) { father[yy]=father[xx]; rank[xx]=rank[xx]+rank[yy]; } else { father[xx]=father[yy]; rank[yy]=rank[xx]+rank[yy]; } }}int main(){ while(cin>>n>>m&&(n||m)) { init(); int k,frist,next; while(m--) { cin>>k>>frist; for(int i=1;i<k;i++) { cin>>next; uniont(frist,next); } } cout<<rank[father[0]]<<endl; } return 0;}
- 并查集poj611suspect
- HDU3938 并查集 并查集
- 并查集(集并查)
- HDU1232 并查集<并>
- 并查集
- 数据结构-并查集
- 并查集
- 并查集!
- 并查集
- 并查集
- 并查集
- 并查集
- 并查集总结
- 并查集学习
- 并查集
- 并查集
- 并查集
- 所谓并查集
- [POJ 1351] Number of Locks
- retrofit资料
- [iOS] iOS开发的22个奇葩技巧
- 1009-I专题四
- linux特殊权限管理
- 并查集poj611suspect
- 【Leetcode】240. Search a 2D Matrix II
- API设计原则总结
- PC能替代服务器吗?全方位解析二者区别
- gcc优化性能简介
- leetCode刷题之路- number344
- # Zxing二维码扫描图片预览变形的问题解决
- 添加IObjectSafety接口使MFC写的OCX可信
- #pragma once与#ifndef两种防止头文件二次编译的区别