Virtual Friends

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 8229 Accepted Submission(s): 2363

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

带权并查集：也就是可以记录每个集合中元素的个数。

实现方法：在普通并查集中增加dist数组在记录每个父节点的深度（即该集合中元素个数）

#include
       
        #include
        
         #include
         #include
          
           using namespace std;#define N 200005map
           
            m;int father[N],dist[N];int Find(int x){    if(father[x]!=x)        father[x]=Find(father[x]);    return father[x];}void Merge(int a,int b){    int fa=Find(a);    int fb=Find(b);    if(fa!=fb)    {        father[fa]=fb;        dist[fb]+=dist[fa];    }    printf("%d\n",dist[fb]);}int main(){    int t;    while(scanf("%d",&t)!=EOF)    {        while(t--)        {            m.clear();            int n,cnt=1;            string name1,name2;            scanf("%d",&n);            for(int i=1; i<=2*n; i++)            {                father[i]=i;                dist[i]=1;            }            while(n--)            {                cin>>name1>>name2;                if(m.find(name1)==m.end())                    m[name1]=cnt++;                if(m.find(name2)==m.end())                    m[name2]=cnt++;                Merge(m[name1],m[name2]);            }        }    }    return 0;}

转载于:https://www.cnblogs.com/jasonlixuetao/p/6115125.html

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