# Let's go to play （模拟）

yanghui07216 · · 3189 次点击 · · 开始浏览

# Let's go to play

### 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.

```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```
`//题意：`
`给你一个n，表示有n个人，接下来输入n行，每行第一个是一个字母，若为M，表示女生，接着一个s（起始天日期），一个e（结束的日期）。现在要求在某一天人数最多，并且男生人数和女生人数相同，问最多人数是多少。`
`直接模拟，然后再从1--366遍历就行了，不能对其排序，排完序后会超时（TML了2次）。`
```#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#define INF 0x3f3f3f3f
#define ull unsigned long long
#define ll long long
#define IN __int64
#define N 380
#define M 1000000007
using namespace std;
struct zz
{
int m;
int f;
int z;
}p[N];
int cmp(zz a,zz b)
{
return a.z<b.z;
}
int main()
{
int t,n,m;
int i,j,k;
char c;
int s,e;
while(scanf("%d",&n)!=EOF)
{
memset(p,0,sizeof(p));
for(i=0;i<n;i++)
{
scanf("%s%d%d",c,&s,&e);
for(j=s;j<=e;j++)
{
if(c=='M')
p[j].m++;
else
p[j].f++;
p[j].z++;
}
}
int mm=0,mi;
for(i=1;i<=366;i++)
{
mi=min(p[i].m,p[i].f);
if(mm<mi)
mm=mi;
}
printf("%d\n",mm*2);
}
return 0;
}``` 0 回复

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# Let's go to play

### 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.

```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```
`//题意：`
`给你一个n，表示有n个人，接下来输入n行，每行第一个是一个字母，若为M，表示女生，接着一个s（起始天日期），一个e（结束的日期）。现在要求在某一天人数最多，并且男生人数和女生人数相同，问最多人数是多少。`
`直接模拟，然后再从1--366遍历就行了，不能对其排序，排完序后会超时（TML了2次）。`
```#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#define INF 0x3f3f3f3f
#define ull unsigned long long
#define ll long long
#define IN __int64
#define N 380
#define M 1000000007
using namespace std;
struct zz
{
int m;
int f;
int z;
}p[N];
int cmp(zz a,zz b)
{
return a.z<b.z;
}
int main()
{
int t,n,m;
int i,j,k;
char c;
int s,e;
while(scanf("%d",&n)!=EOF)
{
memset(p,0,sizeof(p));
for(i=0;i<n;i++)
{
scanf("%s%d%d",c,&s,&e);
for(j=s;j<=e;j++)
{
if(c=='M')
p[j].m++;
else
p[j].f++;
p[j].z++;
}
}
int mm=0,mi;
for(i=1;i<=366;i++)
{
mi=min(p[i].m,p[i].f);
if(mm<mi)
mm=mi;
}
printf("%d\n",mm*2);
}
return 0;
}```