网上交友，计算朋友间都有关系的最多人数，用并查集做，
2，wrong answer的原因：
1、有多组样例
while(scanf("%d",&t)!=EOF) { while(t–) { }
}
2、超时错在用charchange()给出每个人对应的数字编号，改正方法用map;
3/超时解决方式、find方法进行路径压缩、判断0个关系的情况(不确定)
三、题目：
Virtual FriendsTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2085 Accepted Submission(s): 593Problem 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 Input13Fred BarneyBarney BettyBetty Wilma Sample Output234 SourceUniversity of Waterloo Local Contest 2008.09Recommendchenrui
四、代码：
#include<stdio.h>#include<map>#include<iostream>using namespace std;map<string,int> m;int set[100005];int num[100005];int find(int x){int r=x;while(r!=set[r])r=set[r];int i=x;while(i!=r){int j=set[i];set[i]=r;i=j;}return r;}void merge(int x,int y){int fx=find(x);int fy=find(y);if(fx!=fy){set[fx]=fy;num[fy]+=num[fx];printf("%d\n",num[fy]);}else{printf("%d\n",num[fy]);}}int main(){int t;char a[25];char b[25];while(scanf("%d",&t)!=EOF){while(t–){int n;scanf("%d",&n);for(int i=1;i<100005;i++){set[i]=i;num[i]=1;}m.clear();int ans=1;for(int i=1;i<=n;i++){scanf("%s%s",a,b);if(!m[a]){m[a]=ans++;}if(!m[b]){m[b]=ans++;}merge(m[a],m[b]);}}}return 0;}
以一种进取的和明智的方式同它们奋斗。
