Topic 1437:to Fill or not to Fill (greedy algorithm)

Title Description:

With highways available, driving a car from Hangzhou to any other city are easy. But since the tank capacity of a car was limited, we have to find gas stations on the the-on-the-the-the-time. Different give Different price. You is asked to carefully design the cheapest route to go.

input:

for, the first line contains 4 positive Numbers:cmax (<=), the maximum capacity of th e tank; D (<=30000), the distance between Hangzhou and the destination city; Davg (<=20), the average distance per unit gas, the car can run; and N (<=), the total number of gas stations. Then N lines follow, each contains a pair of non-negative Numbers:pi, the unit gas price, and Di (<=d), the distance B Etween This station and the Hangzhou, for I=1,... N. All the numbers in a line is separated by a space.

output:

for each test case, print the cheapest price in a line, accurate up to 2 decimal places. It is assumed, the tank is empty at the beginning. If It is impossible to reach the destination, print "The maximum travel distance = X" where x is the maximum possible dist Ance the car can run, accurate up to 2 decimal places.

Sample input:

50 1300 12 86.00 12507.00 6007.00 1507.10 07.20 2007.50 4007.30 10006.85 30050 1300 12 27.10 07.00 600

Sample output:

749.17The Maximum Travel distance = 1200.00

Analysis:

greedy strategy: Suppose now oneself in a station, to consider is a station to not refueling, add how much oil problem. Locate the current range ( distance from a station CMAX*DAVG) The next station to refuel B.

A station can reach within three kinds of situations:

① There is no gas station,-------to the end, then add the right amount of oil to the end, or impossible, then a station full of oil to what count;

② have cheaper gas stations-------find the first gas station B cheaper than a, add as little oil as possible (or the oil can go straight over) to station B;

③ only a higher-priced petrol station,------a station full of oil, looking for the relatively cheapest gas station B to drive past.

Make a greedy choice for every gas station.

Code:

#include <iostream> #include <stdio.h> #include <algorithm> #include <math.h> #include < Stdlib.h> using namespace std;    struct gas_station{float price;     int distance;    BOOL operator < (const gas_station & A) const {return distance < a.distance;    }}buff[501];int Main () {int cmax,d,davg,n;    int dis; while (scanf ("%d%d%d", &cmax,&d,&davg,&n)!=eof) {for (int i = 0; i < N; i++) {scanf        ("%f%d", &buff[i].price,&buff[i].distance);        } buff[n].distance = D;         Buff[n].price = 100000000;         Sort (buff,buff+n);            if (Buff[0].distance > 0) {printf ("the maximum travel distance = 0.00\n");        Continue dis = Cmax * davg;//maximum single drive distance float sum = 0;//Total cost float Temp,remind_gas = 0;//remaining petrol amount int I, k        ;            for (i = 0; i < n;i++) {k = i+1; if (i! = 0) {Remind_gas-= (float) (buff[i].distance-buff[i-1].distance)/davg;                }//Find the current station after the first price lower than it gas station for (; k < N && buff[i].price <= buff[k].price;k++)            Continue                if (buff[k].distance-buff[i].distance > Dis) {sum + = (cmax-remind_gas) *buff[i].price;            Remind_gas = Cmax;                }else{temp = (float) (buff[k].distance-buff[i].distance)/davg-remind_gas;                    if (fabs (temp) >1e-5&&temp >0) {sum + = Temp*buff[i].price;                Remind_gas = (float) (buff[k].distance-buff[i].distance)/davg; }} if (Buff[i+1].distance-buff[i].distance>dis) {Double total_d = (double) (buf                F[i].distance + dis);                printf ("The maximum travel distance =%.2f\n", total_d);            Break        }} if (I==n) {printf ("%.2f\n", sum);  }} return 0;}

Attention:

Compare floating-point numbers greater than 0 and need to add fabs (temp), because this place also WA once.

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

Topic 1437:to Fill or not to Fill (greedy algorithm)