4-26 旅行规划问题
问题描述
G 先生想独自驾驶汽车从城市 A 到城市 B。从城市 A 到城市 B 的距离为 公里。汽车油箱的容量为 c 公升。每公升汽油能行驶 e 公里。出发点每公升汽油的价格为 p 元。从城市 A 到城市 B 沿途有 n 个加油站。第 i 个加油站距出发点的距离为 ,油价为每公升 元。如 何规划才能使旅行的费用最省。
对于给定的 ,c,e,p,和 n 以及 n 个加油站的距离和油价 和 ,编程计算最小的旅行费用。如果无法到达目的地,输出“No Solution”。
数据输入:
第 1 行是
,c,e,p,和 n。接下来的 n 行中每行 2 个数
和
。
Java
import java.util.Scanner;
public class LvXingGuiHua {
private static double totalDist, capacity, distPerLiter, startPrice;
private static int n;
private static double maxDriveDist, cost;
private static boolean canReach;
private static double MAX = 999999.99;
private static double[] dist,price;
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
while (true) {
totalDist = input.nextDouble();
capacity = input.nextDouble();
distPerLiter = input.nextDouble();
startPrice = input.nextDouble();
n = input.nextInt();
dist = new double[n+2];
price = new double[n+2];
//始点
dist[0] = 0;
price[0] = startPrice;
for(int i=1; i<=n; i++){
dist[i] = input.nextDouble();
price[i] = input.nextDouble();
}
//终点
dist[n+1] = totalDist;
price[n+1] = 0;
maxDriveDist = capacity * distPerLiter;
cost = 0.0;
canReach = true;
for (int i=1; i<n+2; i++)
if(dist[i]-dist[i-1] > maxDriveDist)
canReach = false;
if (!canReach) {
System.out.println("No Solution!");
} else {
driving();
System.out.println(String.format("%.2f",cost));
}
}
}
private static int minPriceStation(int currentStation) {
double minPrice = MAX;
int nextStation = currentStation + 1;
for (int i=currentStation+1; i <=n+1 && (dist[i]-dist[currentStation])<=maxDriveDist; i++) {
if (price[i] < price[currentStation]) {
nextStation = i;
break;
}
if (price[i] < minPrice) {
minPrice = price[i];
nextStation = i;
}
}
return nextStation;
}
private static void driving() {
double rest = 0.0;
double driveDist;
double tmpCost;
int currentStation = 0;
int nextStation;
while (currentStation <= n) {
nextStation = minPriceStation(currentStation);
driveDist = dist[nextStation] - dist[currentStation];
if (price[nextStation] < price[currentStation]) {
tmpCost = (driveDist / distPerLiter - rest) * price[currentStation];
cost += tmpCost;
rest = 0.0;
} else {
tmpCost = (capacity - rest) * price[currentStation];
rest = capacity - driveDist / distPerLiter;
cost += tmpCost;
}
currentStation = nextStation;
}
}
}
Input & Output
275.6 11.9 27.4 2.8 2
102.0 2.9
220.0 2.2
26.95
Reference
王晓东《计算机算法设计与分析》