杭电 Let's go to play

1005 数据已改，第一名的同学，你已经过了，抱歉。。

Let's go to play

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)

Total Submission(s) : 764   Accepted Submission(s) : 213

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Problem Description

Mr.Lin would like to hold a party and invite his friends to this party. He has n friends and each of them can come in a specific range of days of the year from ai to bi.
Mr.Lin wants to arrange a day, he can invite more friends. But he has a strange request that the number of male friends should equal to the number of femal friends.

Input

Multiple sets of test data.

The first line of each input contains a single integer n (1<=n<=5000 )

Then follow n lines. Each line starts with a capital letter 'F' for female and with a capital letter 'M' for male. 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.

Sample Input

4
M 151 307
F 343 352
F 117 145
M 24 128
6
M 128 130
F 128 131
F 131 140
F 131 141
M 131 200
M 140 200

Sample Output

2
4
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MIN(a, b)  (a < b)? a: b
#define MAX_N 6000
using namespace std;
struct node{
    char sex[2];
    int begin,end;
}num[MAX_N];
int main()
{
    int n;
    while (scanf("%d", &n) != EOF) {
        int m, fm;
        for (int i = 0; i < n; i++) {
            scanf("%s%d%d", &num[i].sex, &num[i].begin, &num[i].end);
        }
        int temp = 0;
        for (int i = 1; i < 367; i++) {
            int m = 0, fm = 0;
            for (int j = 0; j < n; j++) {
                if (num[j].begin <= i && num[j].end >= i) {
                    if (num[j].sex[0] == 'M')    m++;
                    else fm++;
                }
            }
            int a = MIN(m, fm);
            if (temp <= a)  temp = a;
        }
        printf("%d
", temp * 2);
    }
    return 0;
}