POJ 2524 :Ubiquitous Religions
2015-12-26 18:00
429 查看
Ubiquitous Religions
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
Sample Output
额。。难道题单错了?。。
。我一看就是并查集。
。所以就这么水过了。。
并查集一A水过
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 23171 | Accepted: 11406 |
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
额。。难道题单错了?。。
。我一看就是并查集。
。所以就这么水过了。。
并查集一A水过
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<vector> #include<queue> #include<sstream> #include<cmath> using namespace std; #define M 100500 int p[M]; int n, m; void start() { for(int i=1; i<=n; i++) p[i] = i; } int find(int x) { return p[x] == x ? x : p[x] = find( p[x] ); } void Kruskal(int x, int y) { int xx = find(x); int yy = find(y); if(xx!=yy) p[yy] = xx; } int main() { int cas = 0; while(scanf("%d%d", &n, &m) &&n &&m) { cas++; start(); int ans = 0; for(int i=1; i<=m; i++) { int x; int y; scanf("%d%d", &x, &y); Kruskal(x, y); } for(int i=1; i<=n; i++) { if( p[i]==i ) ans++; } printf("Case %d: %d\n", cas, ans); } return 0; }
相关文章推荐
- 23种设计模式(4)_创建型_建造者模式(Builder Pattern)
- 【APUE】8、pthread_create函数,创建子线程
- lucene.net 使用过程中的 几个注意事项(含termquery 和QueryParser 的区别)
- iOS8开发之iOS8的UIAlertController
- UITableView 上添加button
- [IOS开发教程] IOS UIDevice & IOS检测屏幕旋转实例
- Educational Codeforces Round 4 C. Replace To Make Regular Bracket Sequence
- LeetCode - N-Queens II
- toString 和String.valueOf
- QueryAddressUtils
- NSString和NSMultableString和NSNumber以及NSValue
- IOS SDK详解之UIAlertController(IOS8之后替代AlertView和ActionSheet)
- LeetCode - N-Queens
- UITextField的属性设置
- IOS UICollectionView 横向分页加载(左滑加载数据)
- UIAlertController 9.0 之后使用
- UISegmentedControl分段控制器
- UIButton的属性设置
- easyui 源码修改,自定义弹窗
- 百度编辑器ueditor每次编辑后多一个空行的解决办法