Codeforces--629B--Far Relative’s Problem（模拟）

Far Relative’s Problem

Time Limit: 2000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u

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Description

Famil Door wants to celebrate his birthday with his friends from Far Far Away. He hasn friends and each of them can come to the party in a specific range of days of the year froma_i tob_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 andb_i (1 ≤ a_i ≤ b_i ≤ 366), providing that thei-th friend can come to the party from daya_i to dayb_i inclusive.

Output

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

Sample Input

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

Sample Output

Hint

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 day140.

n个人在任意一个时间段到达，有男有女，求人最多的时候有多少，并且男等于女

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int a[2][500];int main(){int n;while(scanf("%d",&n)!=EOF){char op[2];int x,y;memset(a,0,sizeof(a));for(int i=0;i<n;i++){scanf("%s%d%d",op,&x,&y);int t=1;if(op[0]=='M') t=0;for(int j=x;j<=y;j++)a[t][j]++;}int ans=0;for(int i=0;i<=366;i++)ans=max(ans,min(a[0][i],a[1][i]));printf("%d\n",ans*2);}return 0;}