H - Horror Film Night

 

Emma and Marcos are two friends who love horror films. This year, and possibly the years hereafter, they want to watch as many films together as possible. Unfortunately, they do not exactly have the same taste in films. So, inevitably, every now and then either Emma or Marcos has to watch a film she or he dislikes. When neither of them likes a film, they will not watch it. To make things fair they thought of the following rule: They can not watch two films in a row which are disliked by the same person. In other words, if one of them does not like the current film, then they are reassured they will like the next one. They open the TV guide and mark their preferred films. They only receive one channel which shows one film per day. Luckily, the TV guide has already been determined for the next 11 million days.

Can you determine the maximal number of films they can watch in a fair way?

Input

The input consists of two lines, one for each person. Each of these lines is of the following form:

• One integer 0≤k≤10000000≤k≤1000000 for the number of films this person likes;

• followed by kk integers indicating all days (numbered by 0,…,9999990,…,999999) with a film this person likes.

Output

Output a single line containing a single integer, the maximal number of films they can watch together in a fair way.

Sample Input 1	Sample Output 1
1 40 2 37 42	3

Sample Input 2	Sample Output 2
1 1 3 1 3 2	2

Sample Input 3	Sample Output 3
1 2 1 2	1

 

Meaning of the questions: There are two friends go to the movies, everyone has their favorite movies;

Rules: (1) two people can not not look like a movie;

           (2) watch two movies at the same time in the same person can not appear like to see the other people do not like to see;

Ideas: All is actually very simple to press a number to re-array, then sort the array, so num = total number, as long as there are two adjacent film with a person like other people do not like it and tot

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
const int maxn =1e6+10;
const  ll mod =1e9+7;
int a[maxn];
int b[maxn];
struct node
{
    int x,y;
}st[maxn*2];
vector<int>p;
vector<int>q;
map<int,int>mp1;
map<int,int>mp2;
bool cmp(node n,node m)
{
    return n.x<m.x;
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        mp1[a[i]]=1;
    }
    int m;
    int num=0;
    scanf("%d",&m);
    int tot=0;
    for(int i=1;i<=m;i++)
    {
        scanf("%d",&b[i]);
        if(mp1[b[i]])
        {
            mp1[b[i]]=0;
            st[++tot].x=b[i];
            st[tot].y=3;
        }
        else
        {
            st[++tot].x=b[i];
            st[tot].y=1;
        }
    }
    for(int i=1;i<=n;i++)
    {
        if(mp1[a[i]])
        {
           st[++tot].x=a[i];
           st[tot].y=2;
        }
    }
    sort(st+1,st+tot+1,cmp);
    num=tot;
    for(int i=1;i<tot;i++)
    {
        if(st[i].y==st[i+1].y&&st[i].y!=3)
        {
            num--;
        }
    }
    cout<<num<<endl;
    return 0;
}

Minus one, the final output can tot. . . . Game, I want complicated, than finish up questions, ac time find themselves too dishes