欢迎使用CSDN-markdown编辑器
来源:互联网 发布:json数据格式 编辑:程序博客网 时间:2024/05/18 18:02
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 9
1 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: 1
Case 2: 7
Hint
Huge input, scanf is recommended.
并查集水题:问题的大概就是找出最多有多少信奉的宗教。。
思路:将同一信仰的合并,然后for一遍,父亲等于本身的累加,所得的结果即为所求
代码如下:
#include<stdio.h>using namespace std;const int maxn=50050;int n;long long m;int pa[maxn];int findfa(int x){ if(pa[x]==x) return x; else return pa[x]=findfa(pa[x]);}void Union(int x,int y){ int xx=findfa(x); int yy=findfa(y); if(xx!=yy) pa[xx]=y;}int main(){ //int n,m; int res=1; while(~scanf("%d%lld",&n,&m)) { if(n==0 && m==0) break; for(int i=0;i<n;i++) { pa[i]=i; } int p,q; for(long long i=0;i<m;i++) { scanf("%d%d",&p,&q); Union(p,q); } int cnt=0; for(int i=0;i<n;i++) if(pa[i]==i) cnt++; printf("Case %d: %d\n",res,cnt); res++; } return 0;}
- 欢迎使用CSDN-markdown编辑器
- 欢迎使用CSDN-markdown编辑器
- 欢迎使用CSDN-markdown编辑器
- 欢迎使用CSDN-markdown编辑器
- 欢迎使用CSDN-markdown编辑器
- 欢迎使用CSDN-markdown编辑器
- 欢迎使用CSDN-markdown编辑器
- 欢迎使用CSDN-markdown编辑器
- 欢迎使用CSDN-markdown编辑器
- 欢迎使用CSDN-markdown编辑器
- 欢迎使用CSDN-markdown编辑器
- 欢迎使用CSDN-markdown编辑器
- 欢迎使用CSDN-markdown编辑器
- 欢迎使用CSDN-markdown编辑器
- 欢迎使用CSDN-markdown编辑器
- 欢迎使用CSDN-markdown编辑器
- 欢迎使用CSDN-markdown编辑器
- 欢迎使用CSDN-markdown编辑器
- matplotlib命令与格式:图例legend语法及设置
- c语言中部分字符串操作函数
- Mybatis Collection查询集合只出现一条数据
- 基于FullCalendar插件的简单个人日程安排系统(1)
- ZigZag Conversion
- 欢迎使用CSDN-markdown编辑器
- javaScript中的if(变量)和if(变量== true)的区别
- PL/SQL学习笔记
- 提升Xcode编译性能--RAM磁盘编译,编译速速飞起来
- Maven項目的創建及SSM環境的搭建
- 多态部分作业 3..创建Rodent(啮齿动物):Mnouse(老鼠),Gerbil(鼹鼠),Hamster(大颊鼠)
- [面试]进程与线程的区别联系,并发和并行的区别
- 线程的同步
- git葵花宝典,GitLabWeb