codeforces 501B B. Misha and Changing Handles

B. Misha and Changing Handles

time limit per test
 1 second

memory limit per test
 256 megabytes

input
 standard input

output
 standard output

Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.

Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.

Input

The first line contains integer q (1 ≤ q ≤ 1000),
the number of handle change requests.

Next q lines contain the descriptions of the requests, one per line.

Each query consists of two non-empty strings old and new,
separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are
distinct. The lengths of the strings do not exceed 20.

The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handlenew is
not used and has not been used by anyone.

Output

In the first line output the integer n — the number of users that changed their handles at least once.

In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old andnew,
separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new.
You may output lines in any order.

Each user who changes the handle must occur exactly once in this description.

Examples

input

5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov

output

3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123

题解：给定一些姓名的原来姓名修改后的名字，由于名字可以修改多次，问最终修改的初始姓名和现在的姓名；由于姓名修改有前后，只能在当前名字上进行修改，列如aa bb，cc aa 这样是不行的，这样就简单了。直接用结构体进行求解。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct st
{
int a;   //标记所用，如果是中间姓名，转换后，标记为0，下次不再访问
char neww[25],old[25];
}num[1005];
int main()
{
int n,t=1;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%s %s",num[i].neww,num[i].old);
num[i].a=i;//只有不为0就好
for(int j=1;j<i;j++)
{
if(num[j].a!=0)
{
if(strcmp(num[j].old,num[i].neww)==0)
{
strcpy(num[j].old,num[i].old);
num[i].a=0; //标记的是当前输入的姓名
}
}
}
}
int ans=0;
for(int i=1;i<=n;i++)
{
if(num[i].a!=0)
ans++;
}
printf("%d\n",ans);
for(int i=1;i<=n;i++)
{
if(num[i].a!=0)
{
printf("%s %s\n",num[i].neww,num[i].old);
}
}
return 0;
}