浙江科技学院第十三届程序设计竞赛1001

Let's go to play

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)
Total Submission(s) : 773   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

<span style="font-family:KaiTi_GB2312;font-size:24px;">#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
int f[370],m[370];
int main()
{
int n,a,b;
char s;
while(scanf("%d",&n)==1)
{
memset(f,0,sizeof(f));
memset(m,0,sizeof(m));

for(int i=0;i<n;i++)
{
getchar();
scanf("%c%d%d",&s,&a,&b);
if(a>b)
{
int t=a;
a=b;
b=a;
}
if(s=='F')
{
for(int j=a;j<=b;j++)
f[j]++;
}
else{
for(int j=a;j<=b;j++)
m[j]++;
}
}
int ans=0;
for(int i=1;i<=366;i++)
{
if(m[i]>ans&&f[i]>ans)
{
ans=m[i]>f[i]?f[i]:m[i];
}
}
printf("%d\n",ans*2);
}
return 0;
}</span>