poj----Ubiquitous Religions
Ubiquitous Religions
Time Limit: 5000MS |
Memory Limit: 65536K |
|
---|---|---|
Total Submissions: 20689 |
Accepted: 10167 |
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 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.
Source
Alberta Collegiate Programming Contest 2003.10.18
代码:
1 #include<iostream>
2 #include<cstdio>
3 #define maxn 50002
4 using namespace std;
5 int Father[maxn],rank[maxn],count;
6 void makeset(int n)
7 {
8 for(int i=1;i<=n;i++)
9 {
10 Father[i]=i;
11 rank[i]=1;
12 }
13 }
14
15 int findset(int x)
16 {
17 if(x!=Father[x])
18 {
19 Father[x]=findset(Father[x]);
20 }
21 return Father[x];
22 }
23
24 void unionset(int fx,int fy)
25 {
26 fx=findset(fx);
27 fy=findset(fy);
28 if(fx==fy)
29 return;
30 if(rank[fx]>rank[fy])
31 {
32 Father[fy]=fx;
33 rank[fx]+=rank[fy];
34 count--;
35 }
36 else
37 {
38 Father[fx]=fy;
39 rank[fy]+=rank[fx];
40 count--;
41 }
42 }
43
44 int main()
45 {
46 int n,m,st,en,num=1;
47 while(scanf("%d%d",&n,&m),n+m)
48 {
49 makeset(n);
50 count=n;
51 while(m--)
52 {
53 scanf("%d%d",&st,&en);
54 unionset(st,en);
55 }
56 printf("Case %d: %dn",num++,count);
57 }
58 //system("Pause");
59 return 0;
60 }
http://poj.org/problem?id=2524
- JavaScript 教程
- JavaScript 编辑工具
- JavaScript 与HTML
- JavaScript 与Java
- JavaScript 数据结构
- JavaScript 基本数据类型
- JavaScript 特殊数据类型
- JavaScript 运算符
- JavaScript typeof 运算符
- JavaScript 表达式
- JavaScript 类型转换
- JavaScript 基本语法
- JavaScript 注释
- Javascript 基本处理流程
- Javascript 选择结构
- Javascript if 语句
- Javascript if 语句的嵌套
- Javascript switch 语句
- Javascript 循环结构
- Javascript 循环结构实例
- Javascript 跳转语句
- Javascript 控制语句总结
- Javascript 函数介绍
- Javascript 函数的定义
- Javascript 函数调用
- Javascript 几种特殊的函数
- JavaScript 内置函数简介
- Javascript eval() 函数
- Javascript isFinite() 函数
- Javascript isNaN() 函数
- parseInt() 与 parseFloat()
- escape() 与 unescape()
- Javascript 字符串介绍
- Javascript length属性
- javascript 字符串函数
- Javascript 日期对象简介
- Javascript 日期对象用途
- Date 对象属性和方法
- Javascript 数组是什么
- Javascript 创建数组
- Javascript 数组赋值与取值
- Javascript 数组属性和方法