hdu 3172 Virtual Friends-白红宇

hdu 3172 Virtual Friends

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发布时间：2019-06-09

本文共 2608 字，大约阅读时间需要 8 分钟。

Problem Description

These days, you can do all sorts of things online. For example, you can use various websites to make virtual friends. For some people, growing their social network (their friends, their friends' friends, their friends' friends' friends, and so on), has become an addictive hobby. Just as some people collect stamps, other people collect virtual friends.

Your task is to observe the interactions on such a website and keep track of the size of each person's network.

Assume that every friendship is mutual. If Fred is Barney's friend, then Barney is also Fred's friend.

Input

Input file contains multiple test cases.

The first line of each case indicates the number of test friendship nest.

each friendship nest begins with a line containing an integer F, the number of friendships formed in this frindship nest, which is no more than 100 000. Each of the following F lines contains the names of two people who have just become friends, separated by a space. A name is a string of 1 to 20 letters (uppercase or lowercase).

Output

Whenever a friendship is formed, print a line containing one integer, the number of people in the social network of the two people who have just become friends.

Sample Input

Fred Barney

Barney Betty

Betty Wilma

Sample Output

解题思路：

此题一道简单的并查集问题。

题目大意是说A与B结成朋友，则此时包含A或者B的朋友圈有多少个人，B与C结成朋友，此时这个朋友圈就有A、B、C三个人，C与D结成朋友，此时这个圈就有4个人了。

所以，将每一个人的名字用map来判断属于第几个出现的名字，将名字转换成数字，然后运用f[]函数，和count[]函数来维护并查集每一个根节点的信息。

代码如下：

1 #include 
        
          2 #include 
         
           3 #include 
           4 #include 
           
             5 #include 
            
              6 using namespace std; 7 int f[111111]; 8 int count[111111]; 9 int t,n;10 11 int find(int a)12 {13     if(a==f[a])14     {15         return a;16     }17     int t=find(f[a]);18     f[a]=t;19     return t;20 }21 int main()22 {23     while(~scanf("%d",&t))24     {25         while(t--)26         {27             scanf("%d",&n);28             string name1,name2;29             map
             
              my_map;30 my_map.clear();31 int i;32 for(i=1;i<=100000;i++)33 {34 f[i]=i;35 count[i]=1;36 }37 38 int cnt=1;39 for(i=0;i
              
               >name1>>name2;42 if(!my_map[name1])43 {44 my_map[name1]=cnt++;45 }46 if(!my_map[name2])47 {48 my_map[name2]=cnt++;49 }50 51 int a=my_map[name1];52 int b=my_map[name2];53 54 int x=find(a);55 int y=find(b);56 57 if(x!=y)58 {59 f[x]=y;60 count[y]+=count[x];61 printf("%d\n",count[y]);62 }63 else64 {65 printf("%d\n",count[x]);66 }67 }68 }69 }70 return 0;71 }

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转载于:https://www.cnblogs.com/ouyangduoduo/archive/2013/05/23/3095760.html

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