Hdu 1350 Taxi Cab Scheme (binary matching)

Hdu 1350 Taxi Cab Scheme (binary matching)

 

 

Taxi Cab SchemeTime Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission (s): 943 Accepted Submission (s): 456

Problem 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 taxi rides which have been booked in advance. given a list of all booked taxi rides for the next day, you want to minimize the number of cabs needed to carry out all of the rides.

For the sake of simplicity, we model a city as a rectangular grid. an address in the city is denoted by two integers: the street and avenue number. the time needed to get from the address a, B to c, d by taxi is | a-c | + | B-d | minutes. A cab may carry out a booked ride if it is its first ride of the day, or if it can get to the source address of the new ride from its latest, at least one minute before the new ride's scheduled departure. note that some rides may end after midnight.

Input On the first line of the input is a single positive integer N, telling the number of test scenarios to follow. each scenario begins with a line containing an integer M, 0 <M <500, being the number of booked taxi rides. the following M lines contain the rides. each ride is described by a departure time on the format hh: mm (ranging from 00:00 to 23:59 ), two integers a B that are the coordinates of the source address and two integers c d that are the coordinates of the destination address. all coordinates are at least 0 and strictly smaller than 200. the booked rides in each scenario are sorted in order of increasing departure time.

Output For each scenario, output one line containing the minimum number of cabs required to carry out all the booked taxi rides.

Sample Input

2208:00 10 11 9 1608:07 9 16 10 11208:00 10 11 9 1608:06 9 16 10 11

Sample Output

12

Source Northwestern Europe 2004
Recommend JGShining, next n rows, each row enters the time, starting position coordinate, and target position coordinate.

 

Output the minimum number of drivers according to the data and requirements given above.

Solution: first, we can see that the first idea of this question is greedy. We can directly compare the arrival time to the start time of the next person. However, we cannot find the minimum number of drivers. In this case, we should use the minimum edge coverage of binary matching, and the minimum path coverage =-the maximum number of matches (n is the number of points)

Specific implementation: Consider the coordinates of each person as a point. If the first driver receives B after receiving a customer, connect a and B to a directed edge. The minimum driver must cover all vertices. Perfect Conversion ~~~

Note:

1. The departure time is Manha distance, formula: | a-c | + | B-d |

2. Each transfer takes one minute, so do not forget to add 1

 

For details, see the code.

 

#include 
 
  #include 
  
   #include 
   
    using namespace std;struct node{    int stime,etime,sx,sy,ex,ey;} p[510];int Map[510][510];int n;int ok[510],vis[510];int fun(int a){    return a<0?-a:a;}bool Find(int x){    for (int i=0; i