Flo's Restaurant[HDU1103]
Flo's Restaurant
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 932 Accepted Submission(s): 285
Problem Description
Sick and tired of pushing paper in the dreary bleary-eyed world of finance, Flo ditched her desk job and built her own restaurant.
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
There are several test cases. The first line of each case contains there positive integers separated by blanks, A, B and C (A, B, C >0, A + B + C <= 100), which are the number of two-seat tables, the number of four-seat tables and the number of six-seat tables respectively. From the second line, there is a list of coming groups of diners, each line of which contains two integers, T and N (0 < N <= 6), representing the arrival time and the number of diners of each group. The arrival time T is denoted by HH:MM, and fixed between 08:00 and 22:00 (the restaurant closes at 23:00). The list is sorted by the arrival time of each group in an ascending order, and you may assume that no groups arrive at the same time. Each test case is ended by a line of “#”.
A test case with A = B = C = 0 ends the input, and should not be processed.
Output
For each test case, you should output an integer, the total number of diners who take their food in Flo’s restaurant, in a separated line.
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
Author
Alcyone
Recommend
Eddy
让排队的时间尽量短
#include<stdio.h>
#include<string.h>
int s[4][125];
int main()
{
while (scanf("%d%d%d",s[1],s[2],s[3])!=EOF)
{
if (s[1][0]+s[2][0]+s[3][0]==0) return 0;
memset(&s[1][1],0,sizeof(s[1])-4);
memset(&s[2][1],0,sizeof(s[2])-4);
memset(&s[3][1],0,sizeof(s[3])-4);
int ans=0,H,M,N,i;
char ch;
getchar();
ch=getchar();
while (ch!='#')
{
scanf("%d:%d%d",&H,&M,&N);
getchar();
H+=(ch-48)*10;
ch=getchar();
int x=(N-1)/2+1,T=H*60+M,v,Min;
bool flag=false;
for (i=1;i<=s[x][0];i++)
{
if (T>=s[x][i])
{
s[x][i]=T+30;
flag=true;
ans+=N;
break;
}
}
if (!flag)
{
Min=2000;
for (i=1;i<=s[x][0];i++)
if (s[x][i]-T<Min)
{
Min=s[x][i]-T;
v=i;
}
if (Min<=30)
{
ans+=N;
s[x][v]+=30;
}
}
}
printf("%d\n",ans);
}
return 0;
}
Flo's Restaurant[HDU1103]的更多相关文章
- HDU 1103 Flo's Restaurant(模拟+优先队列)
Flo's Restaurant Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
- 【模拟】Flo's Restaurant
[poj2424]Flo's Restaurant Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 2960 Accept ...
- 别人整理的DP大全(转)
动态规划 动态规划 容易: , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ...
- [转] POJ DP问题
列表一:经典题目题号:容易: 1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1191,1208, 1276, 13 ...
- poj动态规划列表
[1]POJ 动态规划题目列表 容易: 1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1208, 1276, 13 ...
- POJ动态规划题目列表
列表一:经典题目题号:容易: 1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1191,1208, 1276, 13 ...
- POJ 动态规划题目列表
]POJ 动态规划题目列表 容易: 1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1208, 1276, 1322 ...
- DP题目列表/弟屁专题
声明: 1.这份列表不是我原创的,放到这里便于自己浏览和查找题目. ※最近更新:Poj斜率优化题目 1180,2018,3709 列表一:经典题目题号:容易: 1018, 1050, 1083, 10 ...
- 别人整理的dp题目
动态规划 动态规划 容易: 1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1208, 1276, 1322, 14 ...
随机推荐
- Unity模型导入的若干问题
原地址:http://tieba.baidu.com/p/2807225555 我觉得有三点感觉在导出时比较重要的1.单位的设置3dsmax系统单位设置就设成厘米显示单位公制也设成厘米这样在max里的 ...
- Lucene4.3开发之分词器总结
Lucene4.3开发之分词器总结 http://java.chinaitlab.com/tools/940011.html
- windows 程序的本体与操作系统之间的关系
WinMain(hInst,hPrev,……) { MSG msg; RegisterClass(……); CreateWindow(……); ShowWindow(……); UpdateWindow ...
- Kruskal vs Borůvka
做了个对比.Borůvka算法对于稠密图效果特别好.这两个都是求生成森林的算法.Prim+heap+tarjan过于难写不写了. V=200,E=1000 Kruskal method 4875048 ...
- PHP日期格式转时间戳
PHP 提供了函数可以方便的将各种形式的日期转换为时间戳,该类函数主要是: strtotime():将任何英文文本的日期时间描述解析为时间戳. mktime():从日期取得时间戳. strtotime ...
- Java锁之自旋锁详解
锁作为并发共享数据,保证一致性的工具,在JAVA平台有多种实现(如 synchronized 和 ReentrantLock等等 ) .这些已经写好提供的锁为我们开发提供了便利,但是锁的具体性质以及类 ...
- 算法:comparable比较器的排序原理实现(二叉树中序排序)
Comparable比较器排序远离实现 package test.java.api.api13; /** * 手工实现二叉树的比较算法: 第一遍感觉很神秘,但是真正自己写下来,就感觉很简单,理解就好: ...
- 【JAVA、C++】 LeetCode 008 String to Integer (atoi)
Implement atoi to convert a string to an integer. Hint: Carefully consider all possible input cases. ...
- Java for LeetCode 044 Wildcard Matching
Implement wildcard pattern matching with support for '?' and '*'. '?' Matches any single character. ...
- POJ 2478 Farey Sequence(欧拉函数前n项和)
A - Farey Sequence Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u ...