Virtual Friends

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7434    Accepted Submission(s): 2133


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 1 3 Fred Barney Barney Betty Betty Wilma
Sample Output 2 3 4
Source
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3038 3234 3047 2818 1558  這道題比較難處理的一個地方是需要處理的資料是字串，這裡看別人用了一個STL中的一個對映map是STL中的一個，映照容器，包含在#include<map>中，string是鍵值，int是資料，如果是map["happy"]=6,string str="happy",那麼map[str]=6；

#include<stdio.h>  
#include<map>//注意  
#include<string>  
using namespace std;  
int pre[100005],num[100005];  
map<string,int>m;//STL中的一個映照容器，第一次接觸  
void init()  
{  
    int i;  
    for(i=0;i<=100005;i++)  
    {  
        pre[i]=i;  
        num[i]=1;                 
    }       
}  
int find(int x)  
{  

   if(x==pre[x])
   return x;
   return pre[x]=find(pre[x]); 
}  
void merge(int x,int y)  
{  
    int fx,fy;  
    fx=find(x);  
    fy=find(y);  
    if(fy!=fx)  
    {  
        pre[fx]=fy;  
        num[fy]+=num[fx];  
        printf("%d\n",num[fy]);  
    }  
    else  
    {  
        printf("%d\n",num[fy]);  
    }  
      
}  
int main()  
{  
    int t;  
    while(scanf("%d",&t)!=EOF)//坑   
    {  
          
        while(t--)  
        {  
            int n;  
            char a[200],b[200];  
            init();  
            m.clear(); //清空 
            scanf("%d",&n);  
            int ans=1;  
            while(n--)  
            {  
                scanf("%s %s",a,b);  
                if(!m[a])  
                {  
                    m[a]=ans++;  
                }  
                if(!m[b])  
                {  
                    m[b]=ans++;  
                }  
                merge(m[a],m[b]);  
            }  
        }  
    }  
}