hdu 3172 并查集
Virtual Friends
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3244 Accepted Submission(s): 945
http://acm.hdu.edu.cn/showproblem.php?pid=3172
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.
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
Sample Output
2
3
4
输入格式
while(cin>>t)
{
{
while(t--)
{
...
}
}
map 存储
find 路径压缩
#include<iostream> #include<cstring> #include<cstdio> #include<map> using namespace std; int fa[100010]; int num[100010]; void init() { for(int i=1; i<=100000; i++) { fa[i]=i; num[i]=1; } } int find(int a) { int tmp=a,r=a; while(fa[tmp]!=tmp) tmp=fa[tmp]; while(fa[a]!=tmp) { r=fa[a]; fa[a]=tmp; a=r; } return tmp; } void unon(int a,int b) { int fa1=find(a); int fb1=find(b); if(fa1==fb1) cout<<num[fa1]<<endl; else { num[fa1]+=num[fb1]; fa[fb1]=fa1; cout<<num[fa1]<<endl; } } int main() { int t,n,i,j; char s[25],ss[25]; while(cin>>t) { map<string ,int >p; int mm=1; while(t--) { init(); cin>>n; for(i=1; i<=n; i++) { scanf("%s%s",s,ss); if(!p[s]) p[s]=mm++; if(!p[ss]) p[ss]=mm++; unon(p[s],p[ss]); } } } }