Let's go to play （類比）

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Let's go to play

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)

Total Submission(s) : 771   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

4M 151 307F 343 352F 117 145M 24 1286M 128 130F 128 131F 131 140F 131 141M 131 200M 140 200

Sample Output

24

//題意：

給你一個n，表示有n個人，接下來輸入n行，每行第一個是一個字母，若為M，表示女生，接著一個s（起始天日期），一個e（結束的日期）。現在要求在某一天人數最多，並且男生人數和女生人數相同，問最多人數是多少。

直接類比，然後再從1--366遍曆就行了，不能對其排序，排完序後會逾時（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 1000000007using 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[2];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[0]=='M')p[j].m++;elsep[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;}