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  • 并查集--Ubiquitous Religions

    Ubiquitous Religions

    Time Limit: 1000MS Memory limit: 65536K

    题目描述

    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.

    输入

     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.

    输出

     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.

    示例输入

    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

    示例输出

    Case 1: 1
    Case 2: 7
        #include <stdio.h>
        int bin[100002];
        int findx(int x)
        {
            int r=x;
            while(bin[r] !=r)
                r=bin[r];
            return r;
        }
        void merge(int x,int y)
        {
            int fx,fy;
            fx = findx(x);
            fy = findx(y);
            if(fx != fy)
                bin[fx] = fy;
        }
    
        int main()
        {        int n, m, i, j, x, y, q = 1 , count;//n,m分别为观众数和查询次数,x,y为每次查询观众的编号,q用改变case后的序号
            while(scanf("%d",&n)!=EOF)
            {
                count = 0;
    
                for(i=1;i<=n;i++)
                    bin[i] = i;
                scanf("%d",&m);
                if(n == 0 && m == 0)
                    break;
                for(j=0; j<m; j++)
                {
                    scanf("%d %d",&x,&y);
                    merge(x, y);
                }
              for(i=1; i<=n; i++)
                    if(bin[i] == i)
                    {
                         count ++;
                    }
                printf("Case %d: %d
    ", q++, count);
    
            }
            return 0;
        }
    

    每天训练发现我比别人做的好慢,但是理解的更深刻,如果一开始学一个新知识点就搜模板,那么这样的人是走不远的,毕业之后带走的只有思维,什么荣誉,奖杯都已经不重要了。
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  • 原文地址:https://www.cnblogs.com/6bing/p/3931275.html
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