PAT B -1065 single dog (25 points)

Click on the link full solution summary PAT B -AC

Title:
"Single Dog" is the Chinese term of endearment for a single person. This question you find guests from the single large party of thousands of people in order to give special care.

Input format:
input of the first row is given a positive integer N (≤ 50 000), is the number of known couple / partner; then N rows, each row is given a couple / partner - for convenience, each corresponding to an ID number, digits 5, between the partition ID (from 00,000 to 99,999) by a space; after a given positive integer M (≤ 10 000), the total number of the party; subsequently given the M row guests the ID, separated by a space. Topic ensure no bigamy or foot boats.

Output Format:
First, the single output line of a first total number of guests; followed by a second line ID is incremented by the order listed in the single guests. With a space between the partition ID, and last row may not have extra space.

Sample input:

3
11111 22222
33333 44444
55555 66666
7
55555 44444 10000 88888 22222 11111 23333

Sample output:

5
10000 23333 44444 55555 88888

My code:

#include<iostream>
#include<cstdio>
#include<vector>
#include<string>
#include<set>
#include<map>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<cstring>
#include<sstream>
using namespace std;
//有的时候题目是一起做的，所以会有不需要的头文件

int main()
{
    int pairs[100010]={0};
    int member[100010]={0};
    int N;
    int num_dog=0;
    cin>>N;
    for(int i=0;i<N;i++)
    {
        int a,b;
        cin>>a>>b;
        if(a==0)a=100000;
        if(b==0)b=100000;
        pairs[a]=b;
        pairs[b]=a;
    }
    cin>>N;
    for(int i=0;i<N;i++)
    {
        int a;
        cin>>a;
        if(a==0)a=100000;
        member[a]=1;
    }
    //delete pairs
    for(int i=1;i<100000;i++)
    {
        if(member[i]&&member[pairs[i]])
        {
            member[i]=0;
            member[pairs[i]]=0;
        }
        else if(member[i])
        {
            num_dog++;
        }
    }

    printf("%d\n",num_dog);
    if(member[100000])
    {
        cout<<"00000";
        num_dog--;
        if(num_dog)cout<<" ";
    }
    for(int i=1;i<100000;i++)
    {
        if(member[i])
        {
            printf("%05d",i);
            num_dog--;
            if(num_dog)cout<<" ";
            else break;
        }
    }
    return 0;
}