Ubiquitous Religions(POJ_2524)
2015-08-18 16:54
417 查看
Description
There are so many different religions in the world today that it is difficult to keep track of them all. You are interested in finding out how many different religions students in your university believe in.You know that there are n students in your university (0 < n <= 50000). It is infeasible for you to ask every student their religious beliefs. Furthermore, many students are not comfortable expressing their beliefs. One way to avoid these problems is to ask m (0 <= m <= n(n-1)/2) pairs of students and ask them whether they believe in the same religion (e.g. they may know if they both attend the same church). From this data, you may not know what each person believes in, but you can get an idea of the upper bound of how many different religions can be possibly represented on campus. You may assume that each student subscribes to at most one religion.
Input
The input consists of a number of cases. Each case starts with a line specifying the integers n and m. The next m lines each consists of two integers i and j, specifying that students i and j believe in the same religion. The students are numbered 1 to n. The end of input is specified by a line in which n = m = 0.Output
For each test case, print on a single line the case number (starting with 1) followed by the maximum number of different religions that the students in the university believe in.Sample Input
10 91 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
10 4
2 3
4 5
4 8
5 8
0 0
Sample Output
Case 1: 1Case 2: 7
Hint
Huge input, scanf is recommended.Source
Alberta Collegiate Programming Contest 2003.10.18代码
super简单并查集,没有什么可讲的,直接上代码#include <iostream> using namespace std; int bin[50010]; int find(int x) { return x==bin[x]?x:find(bin[x]); } void merge(int x,int y) { int fx=find(x),fy=find(y); if(fx!=fy) { bin[fx]=fy; } } int main() { int n,m; int a,b; int ans=0; while(cin>>n>>m,n||m) { ans++; for(int i=1;i<=n;i++) bin[i]=i; for(int i=0;i<m;i++) { cin>>a>>b; merge(a,b); } int cnt=0; for(int i=1;i<=n;i++) { if(bin[i]==i) cnt++; } cout<<"Case "<<ans<<": "<<cnt<<endl; } return 0; }
相关文章推荐
- 为 UIImageView 添加Tap手势
- 使用 Fluent API 配置/映射属性和类型
- Arithmetic Sequence
- 取消延迟执行函数 cancelPreviousPerformRequestsWithTarget
- Junit入门学习(2)---------------TestSuite使用
- UIViewController 推出另外一个半透明的UIViewController
- UIScrollView循环滚动图片
- 黑马程序员_GUI
- GPUImage 滤镜头文件
- UIWindow介绍
- OC UIAlertView简化使用
- iOS-UIDynamic-UIKit
- 去哪网实习总结:用到的easyui组件总结(JavaWeb)
- 常识:UI行业常用名词及缩写定义
- JAVA 生成 UUID
- freemaker html页面获取map的key和value
- UITableViewCell drawRect画线在iOS7下不显示的问题
- 让UITableView的headerView或footerView跟随cell一起滚动
- java中StringBuilder、StringBuffer、String类之间的关系
- 排序检索B - List of Conquests