zoukankan      html  css  js  c++  java
  • poj 2524 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

    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
    一开始并不是要做这个题的,而是要做食物链那个题,不过实在想不出具体作法,于是就网搜了,某神牛说做那个题要先做这个题,于是就做了,不过这个题的难度跟那个题没法比啊,就是简单的并查集
    #include<map>
    #include<set>
    #include<queue>
    #include<cmath>
    #include<vector>
    #include<cstdio>
    #include<string>
    #include<cstring>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    #define  inf 0x0f0f0f0f
    
    using namespace std;
    
    const double pi=acos(-1.0);
    const double eps=1e-8;
    typedef pair<int,int>pii;
    
    const int maxn=50001;
    
    int p[maxn];
    bool vis[50001];
    
    int find(int x)
    {
        return x==p[x]?x:p[x]=find(p[x]);
    }
    
    int main()
    {
        //freopen("in.txt","r",stdin);
    
        int n,m,ans,x,y,k=0;
        while (scanf("%d%d",&n,&m)!=EOF && !(n==0 && m==0))
        {
            k++;
            for (int i=1;i<=n;i++) {p[i]=i;vis[i]=false;}
            while (m--)
            {
                scanf("%d%d",&x,&y);
                x=find(x);
                y=find(y);
                p[x]=y;
            }
            ans=0;
            for (int i=1;i<=n;i++) {
                p[i]=find(i);
                if (!vis[p[i]])
                {
                    ans++;
                    vis[p[i]]=true;
                }
            }
            printf("Case %d: %d
    ",k,ans);
        }
        //fclose(stdin);
        return 0;
    }
    至少做到我努力了
  • 相关阅读:
    jquery ui draggable,droppable 学习总结
    VSCode设置网页代码实时预览
    ionic3-修改APP应用图标(icon)和APP启动界面(Splash)
    Ionic3页面的生命周期
    videogular2 在ionic3项目里报错(rxjs_1.fromEvent is not a function)
    IDEA的maven项目的netty包的导入(其他jar同)
    maven的安装与项目的创建
    给定一个排序数组和一个目标值,在数组中找到目标值,并返回其索引。如果目标值不存在于数组中,返回它将会被按顺序插入的位置。
    使用二分法查询二维整型数组的值(找到返回其坐标)
    乐观锁以及悲观锁
  • 原文地址:https://www.cnblogs.com/chensunrise/p/3690089.html
Copyright © 2011-2022 走看看