并查集 HDU 3172 Virtual Friends

题目：

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

除了1个数组记录father，还需要1个数组记录这个集合中的元素数目。

num数组并不是每个数都有意义，只有每个集合的根的num才是这个集合的数目，其他绝大部分num都再也没用了。

（貌似数组开小了，应该扩大1倍，不过还是AC了）

代码：

#include<iostream>
#include<map>
using namespace std;

map<string, int>m;
int fa[100005];
int num[100005];
char s[21];
int k1, k2;
int t, f, k;
map<string, int>::iterator p;

int find(int x)		//找祖先
{
int r = x;
while (fa[r] != r)r = fa[r];
while (fa[x] != x)
{
int  f = fa[x];
fa[x] = r;
x = f;
}
return r;
}

int mapfind(string s)
{
p = m.find(s);
if (p == m.end())
{
m.insert(pair<string, int>(s, k));
fa[k] = k;
num[k++] = 1;
}
return find(m.find(s)->second);
}

int main()
{
while (cin >> t)
{
while (t--)
{
cin >> f;
k = 1;
m.clear();
while (f--)
{
cin >> s;
k1 = mapfind(s);
cin >> s;
k2 = mapfind(s);
if (k1 != k2)
{
fa[k1] = k2;
num[k2] += num[k1];
}
cout << num[k2] << endl;
}
}
}
return 0;
}