Let's go to play

Let's go to play

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)
Total Submission(s) : 770   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
Author

bytelin

区间加法。。。。

详情算法可点击问题 K: 序列的区间操作

代码;

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int male[10050];
int fele[10050];
int fe[10050];
int ma[10050];
int ren[10050];
int main()
{
	char ch[2];
	int n,a,b;
	while (~scanf("%d",&n))
	{
		memset(male,0,sizeof(male));
		memset(fele,0,sizeof(fele));
		memset(ma,0,sizeof(ma));
		memset(fe,0,sizeof(fe));
		for (int i=1;i<=n;i++)	
	    {	
		    scanf("%s%d%d",ch,&a,&b);
		    if (ch[0]=='M')
		    {
			    ma[a]++;ma[b+1]--;
		    }
		    else
		    {
			    fe[a]++;fe[b+1]--;
		    }
	    }
	    for (int i=1;i<=366;i++)
	    {
	    	fele[i]=fele[i-1]+fe[i];
	    	male[i]=male[i-1]+ma[i];
	    	ren[i]=min(fele[i],male[i]);
		}
		ren[0]=0;
		sort(ren,ren+367);
		printf("%d\n",ren[366]*2);
	}
	return 0;
}