UVa10138 - CDVII(排序)

Roman roads are famous for their longevity and sound engineering. Unfortunately, sound engineering does not come cheap, and a number of neo-Caesars have decided to recover the costs through automated tolling.

A particular toll highway, the CDVII, has a fare structure that works as follows: travel on the road costs a certain amount per km travelled, depending on the time of day when the travel begins. Cameras at every entrance and every exit capture the license numbers of all cars entering and leaving. Every calendar month, a bill is sent to the registered owner for each km travelled (at a rate determined by the time of day), plus one dollar per trip, plus a two dollar account charge. Your job is to prepare the bill for one month, given a set of license plate photos.

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

Standard input has two parts: the fare structure, and the license photos. The fare structure consists of a line with 24 non-negative integers denoting the toll (cents/km) from 00:00 - 00:59, the toll from 01:00 - 00:59, and so on for each hour in the day. Each photo record consists of the license number of the vehicle (up to 20 alphanumeric characters), the time and date (mm:dd:hh:mm), the word "enter" or "exit", and the location of the entrance or exit (in km from one end of the highway). All dates will be within a single month. Each "enter" record is paired with the chronologically next record for the same vehicle provided it is an "exit" record. "enter" records that are not paired with an "exit" record are ignored, as are "exit" records not paired with an "enter" record. You may assume that no two records for the same vehicle have the same time. Times are recorded using a 24-hour clock. There are not more than 1000 photo records.

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

Print a line for each vehicle indicating the license number, and the total bill, in alphabetical order by license number. Vehicles that don't use the highway shouldn't be listed.

Sample Input

110 10 10 10 10 10 20 20 20 15 15 15 15 15 15 15 20 30 20 15 15 10 10 10ABCD123 01:01:06:01 enter 17765DEF 01:01:07:00 exit 95ABCD123 01:01:08:03 exit 95765DEF 01:01:05:59 enter 17

Output for Sample Input

765DEF $10.80ABCD123 $18.60

题意：给出一天24小时的收费标准，给出一系列的照片，其信息包含名字，开始时间，是进去还是出来及位置，输出相应的费用

思路：根据名字及开始时间排序

import java.io.FileInputStream;import java.io.BufferedInputStream;import java.io.PrintWriter;import java.io.OutputStreamWriter;import java.util.Scanner;import java.util.Arrays;public class Main implements Runnable{private static final boolean DEBUG = false;private static final int FEE_TYPE = 24;private static int MAXN = 1010;private Scanner cin;private PrintWriter cout;private Photo[] photo = new Photo[MAXN];private int count;private int[] toll = new int[FEE_TYPE];class Photo implements Comparable<Photo>{String license;int time;String status;int location;int fare;public int compareTo(Photo other){if (license.compareTo(other.license) != 0) return license.compareTo(other.license);return time - other.time;}}private void init() {try {if (DEBUG) {cin = new Scanner(new BufferedInputStream(new FileInputStream("d:\\OJ\\uva_in.txt")));} else {cin = new Scanner(new BufferedInputStream(System.in));}cout = new PrintWriter(new OutputStreamWriter(System.out));} catch (Exception e) {e.printStackTrace();}}private void input() {for (int i = 0; i < FEE_TYPE; i++) {toll[i] = cin.nextInt();}count = 0;cin.nextLine();String s;while (true) {if (!cin.hasNextLine()) break;s = cin.nextLine();if (s.length() == 0)break;int current = 0;int pos = s.indexOf(' ', current);photo[count] = new Photo();photo[count].license = s.substring(current, pos);current = pos + 1;pos = s.indexOf(' ', current);String str = s.substring(current, pos);photo[count].time = Integer.parseInt(str.substring(3, 5)) * 24 * 60;photo[count].time += Integer.parseInt(str.substring(6, 8)) * 60;photo[count].time += Integer.parseInt(str.substring(9));current = pos + 1;pos = s.indexOf(' ', current);photo[count].status = s.substring(current, pos);photo[count].location = Integer.parseInt(s.substring(pos + 1));photo[count++].fare = toll[Integer.parseInt(str.substring(6, 8))];}} private void solve(int cas) {Arrays.sort(photo, 0, count);int current = 0;while (current < count){boolean enter = false;int fee = 0;String license = photo[current].license;for (; current < count && photo[current].license.compareTo(license) == 0; current++) {if (!enter) {if (photo[current].status.charAt(1) == 'n') enter = true;else continue;}if (enter) {if (photo[current].status.charAt(1) == 'x') {enter = false;fee += 100;fee += photo[current - 1].fare * Math.abs(photo[current].location - photo[current - 1].location);} else continue;}}if (fee != 0) {fee += 200;cout.print(license);cout.printf(" $%.2f", fee * 1.0 / 100);cout.println();}}if (cas != 0) cout.println();cout.flush();}public void run(){init();int t = cin.nextInt();while (t-- > 0) {input();solve(t);}}public static void main(String[] args) {new Thread(new Main()).start();}}