Codeforces 629B Far Relative’s Problem【模拟】

来源:互联网 发布:淘宝 一千零一夜 编辑:程序博客网 时间:2024/04/30 05:59

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 biinclusive.

Output

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

Examples
input
4M 151 307F 343 352F 117 145M 24 128
output
2
input
6M 128 130F 128 131F 131 140F 131 141M 131 200M 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 345 and 6 can come on day 140.


题目大意:

现在有N个人,有男女之分,要求我们现在选择一天,使得能够来参加聚会的人数最多,其中每个人能够来的时间范围已经给出,并且参加聚会要求一男一女搭伴来,就是说要求男女数量一样。


思路:


1、暴力处理即可,设定m【i】表示第i天有多少个男人可以过来,设定f【i】表示第i天有多少个女人可以过来,对应因为天数在366之内,数量级不大,所以我们直接暴力处理即可。


2、对于其解。暴力扫一遍,取最大值作为解即可。


Ac代码:

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;int m[370];int f[370];int main(){    int n;    while(~scanf("%d",&n))    {        memset(m,0,sizeof(m));        memset(f,0,sizeof(f));        char a[150];        int x,y;        for(int i=1;i<=n;i++)        {            scanf("%s%d%d",&a,&x,&y);            if(a[0]=='M')            {                for(int j=x;j<=y;j++)                {                    m[j]++;                }            }            else            {                for(int j=x;j<=y;j++)                {                    f[j]++;                }            }        }        int output=0;        for(int i=1;i<=368;i++)        {            output=max(output,min(m[i],f[i]));        }        printf("%d\n",output*2);    }}




0 0
原创粉丝点击