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.
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.
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
24
看到这道题一直想着怎么用贪心。。。。看到别人的题解后才恍然大悟,模拟一下就行了。
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int f[50050],m[50050]; int main() { int n,i,j,k,l; while(scanf("%d",&n)!=EOF) { memset(f,0,sizeof(f)); memset(m,0,sizeof(m)); int x,y; char t; for(i=0;i<n;i++) { getchar(); scanf("%c%d%d",&t,&x,&y); if(t=='M') { for(j=x;j<=y;j++) m[j]++; } else { for(j=x;j<=y;j++) f[j]++; } } int ans=0; for(i=1;i<=366;i++) { int minn=min(m[i],f[i]); ans=max(2*minn,ans); } printf("%d\n",ans); } return 0; }
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