Codeforces Round #343 (Div. 2) B

B. Far Relative’s Problem
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
제목:
성별 및 모임에 참가할 수 있는 시간 구간을 입력하다
천사를 선택하면 생일 파티에 참석할 수 있는 사람이 가장 많다(그리고 회의에 참석한 남녀 수가 같다)
문제 풀이:
폭력 매거 1~366일 동안 얼마나 많은 사람들의 구간 주의(성별 구분)를 충족시킬 수 있는지 판단한다ans=max(ans,min(f,m);
 
처음에 할 때는 욕심 구조체 서열로 구간을 하나하나 열거하는 걸 생각했어요.
코드에 원시적인 사고방식의 흔적이 남아 있다

#include<bits/stdc++.h>
using namespace std;
int n;
int f=0,m=0;
int ans=0;
struct node
{
    char sex;
    int l;
    int r;
} N[5005];
bool cmp(struct node a,struct node b)
{
    if(a.l<b.l)
        return true;
    if(a.l==b.l)
    {
        if(a.r>b.r)
            return true;
    }
    return false;
}
int main()
{
    scanf("%d",&n);
    for(int i=0; i<n; i++)
    {
        getchar();
        scanf("%c %d %d",&N[i].sex,&N[i].l,&N[i].r);
    }
    sort(N,N+n,cmp);
    int ans=0;
    for(int i=1; i<=366; i++)
    {
        f=0;
        m=0;
        for(int j=0; j<n; j++)
        {
            if(N[j].l<=i&&i<=N[j].r)
            {
                if(N[j].sex=='F')
                    f++;
                else
                    m++;
            }
        }
        ans=max(ans,min(f,m));
    }
    printf("%d
",ans*2);
    return 0;
}