I - I CodeForces - 629B（暴力区间统计）

I - I CodeForces - 629B（暴力区间统计）

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 a i to b i. 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 a i and b i (1 ≤ a i ≤ b
i ≤ 366), providing that the i-th friend can come to the party from
day a i to day b i 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.

代码

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
void fre() { freopen("A.txt","r",stdin), freopen("Ans.txt","w",stdout); }
#define ll long long 
const int mxn = 400;
const int INF = 0x3f3f3f3f;
struct Node
{
    int sum, f, m;  
} ar[mxn];


int main()
{
    /* fre(); */
    int n;
    scanf("%d", &n);
    char a[5]; int s, e;
    for(int i = 0; i < n; i ++)
    {
        scanf("%s %d %d", a, &s, &e);
        for(int i = s; i <= e; i ++)
        {
            if(a[0] == 'M')
                ar[i].m ++;
            else 
                ar[i].f ++;
        }
    }

    int ans = 0;
    for(int i = 1; i < 367; i ++)
        ans = max(ans, 2 * min(ar[i].m, ar[i].f));
    printf("%d\n", ans);

    return 0;
}