zoukankan      html  css  js  c++  java
  • 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

    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: %d
    ",num++,count);
    57     }
    58     //system("Pause");
    59     return 0;
    60 }
    View Code

    http://poj.org/problem?id=2524

  • 相关阅读:
    外校培训前三节课知识集合纲要(我才不会告诉你我前两节只是单纯的忘了)
    floyd算法----牛栏
    bfs开始--马的遍历
    (DP 线性DP 递推) leetcode 64. Minimum Path Sum
    (DP 线性DP 递推) leetcode 63. Unique Paths II
    (DP 线性DP 递推) leetcode 62. Unique Paths
    (DP 背包) leetcode 198. House Robber
    (贪心 复习) leetcode 1007. Minimum Domino Rotations For Equal Row
    (贪心) leetcode 452. Minimum Number of Arrows to Burst Balloons
    (字符串 栈) leetcode 921. Minimum Add to Make Parentheses Valid
  • 原文地址:https://www.cnblogs.com/gongxijun/p/3264413.html
Copyright © 2011-2022 走看看