【模拟】Flo's Restaurant
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 2960 | Accepted: 929 |
Description
In the small restaurant, there are several two-seat tables, four-seat tables and six-seat tables. A single diner or a group of two diners should be arranged to a two-seat table, a group of three or four diners should be arranged to a four-seat table, and a group of five or six diners should be arranged to a six-seat table.
Flo's restaurant serves delicious food, and many people like to eat here. Every day when lunch time comes, the restaurant is usually full of diners. If there is no suitable table for a new coming group of diners, they have to wait until some suitable table is free and there isn't an earlier arrival group waiting for the same kind of tables. Kind Flo will tell them how long they will get their seat, and if it's longer than half an hour, they will leave for another restaurant.
Now given the list of coming diners in a day, please calculate how many diners take their food in Flo's restaurant. You may assume it takes half an hour for every diner from taking a seat to leaving the restaurant.
Input
A test case with A = B = C = 0 ends the input, and should not be processed.
Output
Sample Input
1 1 1
10:40 1
10:50 2
11:00 4
#
1 1 1
10:40 1
10:50 2
11:00 2
#
1 2 1
10:30 1
10:40 3
10:50 2
11:00 1
11:20 5
#
0 0 0
Sample Output
7
3
12
Source
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
inline int read(){
int x=0,f=1;char c=getchar();
for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
for(;isdigit(c);c=getchar()) x=x*10+c-'0';
return x*f;
}
int A,B,C;
int le1[101],le2[101],le3[101];
int HH,MM,P;
int k[4];
char s[7];
int main(){
k[1]=read(),k[2]=read(),k[3]=read();
while(k[1]!=0||k[2]!=0||k[3]!=0){
int res=0;
memset(le1,0,sizeof(le1));
memset(le2,0,sizeof(le2));
memset(le3,0,sizeof(le3));
while(1){
cin>>s;
if(s[0]=='#') break;
HH=0;
int tmp=1,e=1;
for(tmp=0;s[tmp]!=':';tmp++) ;
for(int i=tmp-1;i>=0;i--)
HH+=((s[i]-'0')*e),e*=10;
MM=0,e=1;
for(int i=strlen(s)-1;i>tmp;i--) MM+=((s[i]-'0')*e),e*=10;
int T=(HH-7)*60+MM;
//cout<<T<<endl;
P=read();
P++;
if(P/2==1&&k[1]){
P--;
sort(le1,le1+k[1]);
if(le1[0]-T<=30){
// cout<<"true"<<endl;
res+=P;
if(le1[0]>=T) le1[0]+=30;
else le1[0]=T+30;
}
}
else if(P/2==2&&k[2]){
P--;
sort(le2,le2+k[2]);
if(le2[0]-T<=30){
//cout<<"true2"<<endl;
res+=P;
if(le2[0]>=T) le2[0]+=30;
else le2[0]=T+30;
}
}
else if(P/2==3&&k[3]){
P--;
sort(le3,le3+k[3]);
if(le3[0]-T<=30){
//cout<<"true3"<<endl;
res+=P;
if(le3[0]>=T) le3[0]+=30;
else le3[0]=T+30;
}
}
}
printf("%d\n",res);
k[1]=read(),k[2]=read(),k[3]=read();
}
}
/*
1 3 2
09:45 3
09:46 5
09:59 4
10:00 1
10:01 3
10:20 3
10:30 2
11:20 5
11:40 1
12:00 3
12:10 2
14:20 5
15:00 3
15:20 6
15:40 5
17:30 2
19:00 6
19:40 4
20:03 4
20:05 4
21:40 2
21:44 1
22:00 1
#
*/
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