codefoces-Far Relative’s Problem

题目
B. Far Relative’s Problem
Famil Door wants to celebrate his birthday with his friends from Far Far Away. He has n friends and each of them can come to the party in a specific range of days of the year from ai to bi. Of course, Famil Door wants to have as many friends celebrating together with him as possible.

Far cars are as weird as Far Far Away citizens, so they can only carry two people of opposite gender, that is exactly one male and one female. However, Far is so far from here that no other transportation may be used to get to the party.

Famil Door should select some day of the year and invite some of his friends, such that they all are available at this moment and the number of male friends invited is equal to the number of female friends invited. Find the maximum number of friends that may present at the party.

Input
The first line of the input contains a single integer n (1 ≤ n ≤ 5000) — then number of Famil Door’s friends.

Then follow n lines, that describe the friends. Each line starts with a capital letter ‘F’ for female friends and with a capital letter ‘M’ for male friends. Then follow two integers ai and bi (1 ≤ ai ≤ bi ≤ 366), providing that the i-th friend can come to the party from day ai to day bi inclusive.

Output
Print the maximum number of people that may come to Famil Door’s party.

Examples
input
4
M 151 307
F 343 352
F 117 145
M 24 128
output
2
input
6
M 128 130
F 128 131
F 131 140
F 131 141
M 131 200
M 140 200
output
4
Note
In the first sample, friends 3 and 4 can come on any day in range [117, 128].

In the second sample, friends with indices 3, 4, 5 and 6 can come on day 140.

（简单思路题）
**题意：
**给你一组数据，数据中M表示男生，F表示女生，每个字母后面跟两个数字，第一个数字是表示这个人可以从这一天（包含这一天）去参加聚会，第二个数字表示到这一天（也包含这一天）结束。
首先知道，去参加的聚会的人必须是一对，问你找到一个时间（某一天）找出能参加的聚会的最多的对，并输出参加人数；

解决：暴力；

我们可以把每个F和M的每个时间段分别都标记出现次数，然后找出它们同一时间段里标记出现小的那个就是可以组成对的个数，然后乘与2就是人数，依次找即可；

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#define ll long long
using namespace std;
const int num=50005;
int b[num];
int c[num];
int main()
{
    int t,x,y;
    cin>>t;
    getchar();
    char a;
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    for(int i=1;i<=t;i++)
    {
        cin>>a>>x>>y;
        if(a=='M')
        for(int j=x;j<=y;j++)
        {
            b[j]++;
        }
        else if(a=='F')
        {
            for(int j=x;j<=y;j++)
            {
                c[j]++;
            }
        }
    }
    int sum=0;
    for(int i=1;i<=5000;i++)
    {
        if(sum<(min(b[i],c[i])*2))
        sum=(min(b[i],c[i])*2);
    }
    cout<<sum;
    return 0;
}