Taxi Cab Scheme 时间限制: 1 Sec 内存限制: 64 MB 题目描述 Running a taxi station is not all that simple. Apart from the obvious demand for a centralised coordination of the cabs in order to pick up the customers calling to get a cab as soon as possible,there is…
Running a taxi station is not all that simple. Apart from the obvious demand for a centralised coordination of the cabs in order to pick up the customers calling to get a cab as soon as possible,there is also a need to schedule all the taxi rides whi…
http://acm.hdu.edu.cn/showproblem.php?pid=1151 Air Raid Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 7469 Accepted: 4451 Description Consider a town where all the streets are one-way and each street leads from one intersection to an…
The Global Aerial Research Centre has been allotted the task of building the fifth generation of mobile phone nets in Sweden. The most striking reason why they got the job, is their discovery of a new, highly noise resistant, antenna. It is called 4D…
Description Running a taxi station is not all that simple. Apart from the obvious demand for a centralised coordination of the cabs in order to pick up the customers calling to get a cab as soon as possible,there is also a need to schedule all the ta…
Taxi Cab Scheme Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 231 Accepted Submission(s): 142 Problem Description Running a taxi station is not all that simple. Apart from the obvious deman…
http://poj.org/problem?id=2060 Taxi Cab Scheme Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 5459 Accepted: 2286 Description Running a taxi station is not all that simple. Apart from the obvious demand for a centralised coordination…
题意:有 n 个客人,要从 si 到 ti,每个人有一个出发时间,现在让你安排最少和出租车去接,在接客人时至少要提前一分钟到达客人的出发地点. 析:把每个客人看成一个结点,然后如果用同一个出租车接的话,那么肯定是先接 u 然后再去接 v,也就是有一条边 u->v,画图看的就成知道,这是一个最小路径覆盖的问题.把每个结点拆成 X和 Y 然后如果 u 能连 v,那么就 Xu -> Yv,然后跑一次二分最大匹配,那么答案就是 n - 最大匹配数. 代码如下: #pragma comment(link…
题意: 出租车公司有M个订单. 订单格式: hh:mm a b c d 含义:在hh:mm这个时刻客人将从(a,b)这个位置出发,他(她)要去(c,d)这个位置. 规定1:从(a,b)到(c,d)所花的时间为:abs(a-c)+abs(b-d). 规定2:一辆出租车如果要接单,必须在客人出发前1分钟(包括)以前接单. 问,最少派出多少辆出租车,可以完成所有任务. 思路: 把每一笔单看成一个点,如果完成第i个单后可以接到第j个单,则 i->j连上线. 则题为:求这个DAG图的最小路…
传送门:http://poj.org/problem?id=2060 Taxi Cab Scheme Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 7260 Accepted: 2972 Description Running a taxi station is not all that simple. Apart from the obvious demand for a centralised coordination…
Taxi Cab Scheme Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 712 Accepted Submission(s): 337 Problem Description Running a taxi station is not all that simple. Apart from the obvious demand f…
