zoukankan      html  css  js  c++  java
  • POJ 2524 :Ubiquitous Religions

    Time Limit: 5000MS
    Memory Limit: 65536K
    Total Submissions: 23171
    Accepted: 11406

    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
    


    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 ", cas, ans); } return 0; }







  • 相关阅读:
    python列表转json树菜单
    lvm分区创建和扩容
    分布式网络概述
    mycat权威指南阅读笔记--序言1
    Mongodb副本集实现及读写分离
    线程和进程
    socket客户端怎么判断http响应数据的结束
    java遍历http请求request的所有参数实现方法
    Java中mongodb使用and和or的复合查询
    idea @Override is not allowed when implementing interface method
  • 原文地址:https://www.cnblogs.com/mfrbuaa/p/5078519.html
Copyright © 2011-2022 走看看