IEEEXtreme 10.0 - Food Truck

这是 meelo 原创的 IEEEXtreme极限编程大赛题解

 

Xtreme10.0 - Food Truck

题目来源 第10届IEEE极限编程大赛

https://www.hackerrank.com/contests/ieeextreme-challenges/challenges/food-truck

 

Madhu has a food-truck called "The Yummy Goods" that goes to a different business hotspot every day at lunch! Madhu wants to perform location-based advertising to folks in the offices near her halt. To do this she uses the GPS location as a longitude and a latitude at the stop and decides on a radius (r) value. She wants to broadcast advertisement SMSes, to customers within this radius, advertising her food-truck.

She needs your help to generate the list of phone numbers of such folks. She has access to a big file of telecom data, which among other details, contains the phone number, longitude and latitude of active cell-phone users in the city at that moment.

In order to calculate the distance between her stops and her subscribers, she wants you to use the most recent location available for each subscriber. To calculate the distance, you should use the Haversine formula:

d = 2 × r × arcsin (sqrt (sin2((lat1 - lat2)/2) + cos(lat1) × cos(lat2) × sin2((long1 - long2)/2)))

where d is the distance between two points on the surface of the earth, in km's

r is the radius of the earth (6378.137 km for this problem)

lat1, long1 are the latitude and longitude, respectively, of point 1

lat2, long2 are the latitude and longitude, respectively, of point 2

Input Format

The first line contains Madhu's latitude and longitude in degrees, separated by a comma.

The second line contains the radius r in kms, within which she wants to broadcast her advertisement.

The third line is a header for the data in the subsequent lines.

The remaining lines have rows of telecom data of active cellphone users. Each line contains the following comma-separated fields:

• A time stamp in MM/DD/YYYY hh:mm format. MM, is a two-digit month, e.g. 01 for January, DD is a two-digit day of month (01 through 31), YYYY is a four-digit year, hh is the two digits of hour (00 through 23), and mm is the two digits of minute (00 through 59)

• The latitude of the subscriber, in degrees

• The longitude of the subscriber, in degrees

• The subscriber's phone number, as a 10-digit number

Notes:

• Some subscribers may appear multiple times. You should use the most recent entry to determine the location and phone number of a subscriber. If a subscriber appears multiple times, the date/time stamps will differ.

• None of the field values will contain commas.

Constraints

In order to eliminate rounding and approximation errors, no subscribers will be at a distance d from Madhu, such that 0.99 × r ≤ d ≤ 1.01 × r

1 ≤ r ≤ 100

There will be at most 50,000 lines in the subscriber list.

Output Format

A comma separated list of phone numbers for subscribers within a radius r of the stop, sorted in ascending order.

Sample Input

 

18.9778972,72.8321983
1.0
Date&Time,Latitude,Longitude,PhoneNumber
10/21/2016 13:34,18.912875,72.822318,9020320100
10/21/2016 10:35,18.9582233,72.8275845,9020320024
10/21/2016 15:20,18.95169982,72.83525604,9020320047
10/21/2016 15:23,18.9513048,72.8343388,9020357980
10/21/2016 15:23,18.9513048,72.8343388,9020357962
10/21/2016 15:28,18.9548652,72.8332443,9020320027
10/21/2016 14:03,18.9179784,72.8279306,9020357972
10/21/2016 14:03,18.9179784,72.8279306,9020357959
10/21/2016 09:52,18.97523123,72.83494895,9020320007
10/21/2016 09:44,18.9715932,72.8383992,9020357607
10/21/2016 09:44,18.9715932,72.8383992,9020357593
10/21/2016 09:44,18.9715932,72.8383992,9020357584
10/21/2016 14:57,18.93438826,72.82704499,9020320011
10/21/2016 09:56,18.97596514,72.8327072,9020320045
10/21/2016 08:33,18.9811929,72.8353202,9020320084
10/21/2016 13:27,18.9159265,72.8245989,9020357896
10/21/2016 13:09,18.9077347,72.8076201,9020320094
10/21/2016 10:52,18.97523003,72.83494865,9020320007

 

Sample Output

9020320007,9020320045,9020320084,9020357584,9020357593,9020357607

Explanation

We can calculate the distance between the location "18.9778972, 72.8321983" and each of the subscribers whose details are provided. Only the 6 phone numbers, listed in the Sample Output set, have a distance to the location of the food-truck that is less than 1.0 km.

 

题目解析

题目本身并没什么难的，但有几个地方容易犯错误。

几个注意事项：

• 首先Haversine公式里的经纬度是弧度、而输入的数据单位是角度，需要进行转换。
• 不要忘了所有的经纬度都要转换，包括店铺的位置和客户的位置。
• 读入数据用scanf比cin要容易一些。
• 可以创建一个结构体来保存用户的数据，key为电话号码。
• 输出的升序是电话号码的升序。
• 读取行数不确定的输入，可以用while(scanf(...) != EOF)
• 比较时间的前后关系，可以先把时间转换成一个数，转换过程中可以把每个月都算31天，每年都算366天。

 

程序

C++

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <map>
#include <string>

#define PI 3.1415926

using namespace std;

struct Custom {
    int year, month, day, hour, minute, second;
    double lat, long_;
    
    int timeNumber() const {
        return second + minute*60 + hour*3600 + day*86400 
               + month*2678400 + (year-1970)*980294400;
    }
    
    bool earlyThan(const Custom & that) {
        return timeNumber() < that.timeNumber();
    }
    
    bool inRange(double r, double myLat, double myLong) const {
        double d = 2 * 6378.137 * asin (sqrt ( pow(sin((lat-myLat)/2),2) + cos(lat) * cos(myLat) * pow(sin((long_-myLong)/2),2)));
        return d < r;
    }
};

double degree2radian(double degree) {
    return degree / 180 * PI;
}

int main() {
    double myLat, myLong, r;
    
    scanf("%lf,%lf", &myLat, &myLong);
    myLat = degree2radian(myLat);
    myLong = degree2radian(myLong);
    
    scanf("%lf", &r);
    map<long long, Custom> customs;

    string ignore;
    cin >> ignore;

    Custom c;
    long long phone;
    while(scanf("%d/%d/%d %d:%d,%lf,%lf,%lld", &c.day, &c.month, &c.year, &c.hour, &c.minute, &c.lat, &c.long_, &phone) != EOF) {
        c.lat = degree2radian(c.lat);
        c.long_ = degree2radian(c.long_);
        if(customs.find(phone) == customs.end()) {
            customs[phone] = c;
        }
        else {
            if(customs[phone].earlyThan(c)) {
                customs[phone] = c;
            }
        }
        c = customs[phone];
    }

    bool first = true;
    for(const auto kv : customs) {
        if( kv.second.inRange(r, myLat, myLong) ) {
            if(!first) cout << ',';
            first = false;
            cout << kv.first;
        }
    }

    return 0;
}

 Python2

from datetime import datetime
from math import radians, cos, sin, asin, sqrt
import sys

def haversine(lon1, lat1, lon2, lat2):
    """
    Calculate the great circle distance between two points 
    on the earth (specified in decimal degrees)
    """
    # convert decimal degrees to radians 
    lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])

    # haversine formula 
    dlon = lon2 - lon1 
    dlat = lat2 - lat1 
    a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
    c = 2 * asin(sqrt(a)) 
    r = 6378.137
    return c * r

lat, lon = map(float, raw_input().split(","))
r = float(raw_input())
raw_input()
d = {}
d = {}
for line in sys.stdin:
    subscriber  = line.strip().split(",")

    dat = datetime.strptime(subscriber[0], "%m/%d/%Y %H:%M")
    h = haversine(lon, lat, float(subscriber[2]), float(subscriber[1]))
    if subscriber[3] in d:
        if dat > d[subscriber[3]][0]:
            d[subscriber[3]] = [dat, h]
    else:
        d[subscriber[3]] = [dat, h]
res = []
for phone in d:
    if  d[phone][1]<=r:
        res.append(phone)

print ",".join([phone for phone in sorted(res)])

from: blog.chaker.tn/posts/food-truck-ieeextreme-10-0

 

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