*HDU3172 并查集

Virtual Friends

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8165    Accepted Submission(s): 2345

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
 
Source
 
题意:
朋友的朋友是朋友,每输入一对朋友后问一共有几对朋友。
代码:
 //用map把字符串转化为数字。初始化时初始2*n个会超时。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
using namespace std;
int t,n;
int num[],fat[];
map<string,int>m;
int find(int x)
{
int rt=x;
while(fat[rt]!=rt)
{
rt=fat[rt];
}
int i=x,j;
while(i!=rt)
{
j=fat[i];
fat[i]=rt;
i=j;
}
return rt;
}
void connect(int x,int y)
{
int xx=find(x),yy=find(y);
if(xx!=yy)
{
fat[xx]=yy;
num[yy]+=num[xx];
printf("%d\n",num[yy]);
}
else printf("%d\n",num[yy]);
}
int main()
{
char ch1[],ch2[];
while(scanf("%d",&t)!=EOF)
{
while(t--)
{
scanf("%d%d",&n);
for(int i=;i<=;i++) //开大了会超时
{
fat[i]=i;
num[i]=;
}
m.clear();
int cnt=;
for(int i=;i<=n;i++)
{
scanf("%s%s",ch1,ch2);
if(!m[ch1]) m[ch1]=cnt++;
if(!m[ch2]) m[ch2]=cnt++;
connect(m[ch1],m[ch2]);
}
}
}
return ;
}
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