标签：color 不同 return str1 href obb map case while
题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=3172
Virtual Friends
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9227 Accepted Submission(s): 2669Problem DescriptionThese 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.
InputInput 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).
OutputWhenever 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 Input1 3 Fred Barney Barney Betty Betty Wilma
Sample Output2 3 4
Source
#include<cstdio> #include<iostream> #include<map> #include<cstring> using namespace std; const int maxn= 110000; int pre[maxn],num[maxn]; char name1[30],name2[30]; string str1,str2; map<string,int> mmap; int find(int x) { int r=x; while(r!=pre[r]) r=pre[r]; int i=x,j; while(pre[i]!=r) { j=pre[i]; pre[i]=r; i=j; } return r; } void join(int x,int y) { int fx=find(x),fy=find(y); if(fx!=fy) { pre[fx]=fy; num[fy]+=num[fx]; num[fx]=0; } printf("%d\n",num[fy]); ///每次都要输出 } int main() { int t,n; while(~scanf("%d",&t)) ///多组数据 { while(t--) { mmap.clear(); scanf("%d",&n); for(int i=1; i<maxn; i++) pre[i]=i,num[i]=1; int no=1; while(n--) { scanf("%s%s",name1,name2); str1=name1; str2=name2; if(mmap.find(str1)==mmap.end()&&mmap.find(str2)!=mmap.end()) { mmap[str1]=no++; join(mmap[str1],mmap[str2]); } else if(mmap.find(str2)==mmap.end()&&mmap.find(str1)!=mmap.end()) { mmap[str2]=no++; join(mmap[str1],mmap[str2]); } else if(mmap.find(str2)==mmap.end()&&mmap.find(str1)==mmap.end()) { mmap[str1]=no++; mmap[str2]=no++; join(mmap[str1],mmap[str2]); } else if(mmap.find(str2)!=mmap.end()&&mmap.find(str1)!=mmap.end()) { join(mmap[str1],mmap[str2]); } } } } return 0; }
标签：color 不同 return str1 href obb map case while
原文地址：http://www.cnblogs.com/longl/p/7281254.html
