HDU Virtual Friends（超级经典的带权并查集）

Virtual Friends

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11092    Accepted Submission(s): 3221

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

University of Waterloo Local Contest 2008.09

 

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题目说的是，网上交友，朋友的朋友也是自己的朋友，问每次两个交了朋友之后，他们朋友圈里有多少人，

并查集处理比较顺，用上map 编号，比较简单就处理了.....但是这个题目的输入比较坑，很少见到这样的输入..........无限次指定次数的输入.........

分析：

#include<queue>
#include<set>
#include<cstdio>
#include <iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<map>
#include<string>
#include<string.h>
#include<memory>
using namespace std;
#define max_v 100005
#define INF 9999999
int pa[max_v+5];
int num[max_v+5];
int n;
void init()
{
    for(int i=0; i<max_v; i++)
    {
        pa[i]=i;
        num[i]=1;//权
    }
}
int find_set(int x)
{
    if(pa[x]!=x)
        pa[x]=find_set(pa[x]);
    return pa[x];
}
void union_set(int x,int y)
{
    int fx=find_set(x);
    int fy=find_set(y);

    if(fx!=fy)
    {
        pa[fx]=fy;
        num[fy]+=num[fx];//！！！计数
    }
    printf("%d
",num[fy]);
}
int main()
{
    int t;
    while(~ scanf("%d",&t))
    {
        string str1,str2;
        while(t--)
        {
            init();
            map<string,int>mm;
            scanf("%d",&n);
            int k=1;
            for(int i=0; i<n; i++)
            {
                cin>>str1>>str2;
                if(!mm[str1])
                {
                    mm[str1]=k++;
                }
                if(!mm[str2])
                {
                    mm[str2]=k++;
                }
                union_set(mm[str1],mm[str2]);
            }
        }
    }
    return 0;
}