HDU3172 Virtual Friends 並查集

 

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

 

13Fred BarneyBarney BettyBetty Wilma

 

 

Sample Output

 

234

 

   題意是這樣的：給你兩個人，他們是朋友關係，判斷在他們朋友關係的集合中有幾對這樣的關係，比如說小明和小紅是朋友，小紅和小李是朋友，那麼小明和小李也是朋友；題意很明白就是並查集的簡單應用，關鍵是名字如何用數字來一一表示。這裡用map來實現，還有很多實現方法，我願聞其詳。

代碼：

#include<iostream>#include<string>using namespace std;#define MAX 100005int parent[MAX],deep[MAX];#include<map>int Max;map <string,int> A;int Find(int x){int s=x;int temp;while(s!=parent[s]){s=parent[s];}while(s!=x){temp=parent[x];parent[x]=s;x=temp;}return s;}void Union(int x,int y){int a=Find(x);int b=Find(y);if(a==b)return;if(a!=b){parent[a]=b;deep[b]+=deep[a];}}int main(){int T,n;string name1,name2;while(scanf("%d",&T)!=EOF){while(T--){scanf("%d",&n);Max=0;A.clear();while(n--){cin>>name1>>name2;if(A.find(name1)==A.end()){Max++;A[name1]=Max;parent[Max]=Max;deep[Max]=1;}if(A.find(name2)==A.end()){Max++;A[name2]=Max;parent[Max]=Max;deep[Max]=1;}Union(A[name1],A[name2]);int p=Find(A[name2]);printf("%d\n",deep[p]);}    }}return 0;}

求效率更好的實現方法。