Step5.1.7virtual friends,3172
VirtualFriends
Time Limit: 4000/2000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 230 AcceptedSubmission(s): 77
Problem Description
These days, you can do all sorts of thingsonline. For example, you can use various websites to make virtual friends. Forsome people, growing their social network (their friends, their friends'friends, their friends' friends' friends, and so on), has become an addictivehobby. Just as some people collect stamps, other people collect virtualfriends.
Your task is to observe the interactions onsuch a website and keep track of the size of each person's network.
Assume that every friendship is mutual. IfFred 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 thenumber of test friendship nest.
each friendship nest begins with a linecontaining an integer F, the number of friendships formed in this frindshipnest, which is no more than 100 000. Each of the following F lines contains thenames of two people who have just become friends, separated by a space. A nameis a string of 1 to 20 letters (uppercase or lowercase).
Output
Whenever a friendship is formed, print aline containing one integer, the number of people in the social network of thetwo 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 LocalContest 2008.09
Recommend
chenrui
題解：
並查集水題，這道題是要求如果是朋友就併到一起，每次輸入朋友，則對應輸出朋友圈的個數。簡單的並查集，當兩個人不在同一個朋友圈的時候，將兩個朋友合併，就是讓小的朋友圈的根改爲大的根，其中根記錄着整個朋友圈的人數，將根相加就可以了。
源代碼：
#include <iostream>
#include <string>
#include <map>
#include <stdio.h>
using namespace std;
struct node
{
intparents;
intnum;
}fri[200005];
map<string,int> name;
void init(int n)
{
n= n * 2 + 1;
for(inti = 0;i < n;i++)
{
fri[i].parents= i;
fri[i].num= 1;
}
}
int find_parent(int x)
{
if(x== fri[x].parents)
returnx;
returnfri[x].parents = find_parent(fri[x].parents);
}
int Union(inta,int b)
{
intx = find_parent(a);
inty = find_parent(b);
if(x== y)
{
intk = fri[x].parents;
returnfri[k].num;
}
if(fri[x].num> fri[y].num)
{
fri[y].parents= x;
fri[x].num+= fri[y].num;
returnfri[x].num;
}
else
{
fri[x].parents= y;
fri[y].num+= fri[x].num;
returnfri[y].num;
}
}
int main()
{
intn,m;
while(cin>> n)
{
for(intx = 0; x < n;x++)
{
scanf("%d",&m);
init(m);
charname1[25],name2[25];
intit = 0,i1,i2;
for(inti = 0;i < m;i++)
{
scanf("%s%s",name1,name2);
if(name.find(name1)== name.end())
{
name[name1]= it;
i1= it++;
}
else
i1= name[name1];
if(name.find(name2)== name.end())
{
name[name2]= it;
i2= it++;
}
else
i2= name[name2];
printf("%d\n",Union(i1,i2));
}
name.clear();
}
}
return0;
}