Virtual Friends
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5491 Accepted Submission(s): 1519
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).
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
3
Fred Barney
Barney Betty
Betty Wilma
Sample Output
2
3
4
Source
题意： 逐步给你一些关系网，对于每一步求所给出的两个人合并之后所构成的关系网。
代码：
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<map>
#include<string>
using namespace std;
const int maxn=;
int father[maxn];
int rank[maxn];
//初始化
void init( int n)
{
for(int i =;i<=n ;i++){
father[i]=i;
rank[i]=;
}
}
//搜索
int fin(int x)
{
int tem=x;
while(x!=father[x]){
x=father[x];
}
//进一步压缩
while(tem!=father[tem])
{
tem=father[tem];
father[tem]=x;
} return x;
}
void Union(int a,int b){
a=fin(a);
b=fin(b);
if(a!=b){
if(rank[a]<rank[b]){
rank[b]+=rank[a];
father[a]=b;
}
else{
rank[a]+=rank[b];
father[b]=a;
}
}
}
map<string,int>sac;
char aa[maxn][],bb[maxn][]; int main()
{
int t,n;
while(scanf("%d",&t)!=EOF)
{
while(t--){
scanf("%d",&n);
if(!sac.empty()) sac.clear();
int cnt=;
for(int i=;i<n ;i++){
scanf("%s%s",aa[i],bb[i]);
// map<string,int>::iterator it;
//it=sac.find(aa[i]);
if(sac.find(aa[i])==sac.end())
sac[aa[i]]=++cnt; // it=sac.find(bb[i]);
if(sac.find(bb[i])==sac.end())
{
// posb=sac.size();
// sac.insert(pair<string,int>(bb,posb));
sac[bb[i]]=++cnt;
}
}
init(cnt);
for(int i=;i<n;i++)
{
Union(sac[aa[i]],sac[bb[i]]);
printf("%d\n",rank[fin(sac[aa[i]])]);
}
}
}
return ;
}