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

 

题意：有两个朋友 一起去看电影，每个人都有自己喜欢的电影；

规则：（1）两个人不能看都不喜欢的电影；

           （2）看的两场电影中不能出现同时为同一个人喜欢看，而另一个人不喜欢看；

思路：其实很简单就是将所有数压到一个数组中去重，然后对这个数组排序，让num=总数，只要出现两个相邻的电影同为一个人喜欢且另一个人不喜欢就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;
}

减一，最后输出tot就可以。。。。比赛的时候，想复杂了，比完赛补题，ac的时候发现自己太菜了

 

来源：CSDN

作者：Ramzes_666_fan

链接：https://blog.csdn.net/Ramzes_666_fan/article/details/103952152