网易机试-赶去公司_去公司机试

输入描述:

输入数据包括五行:

第一行为周围出租车打车点的个数n(1 ≤ n ≤ 50)

第二行为每个出租车打车点的横坐标tX[i] (-10000 ≤ tX[i] ≤ 10000)

第三行为每个出租车打车点的纵坐标tY[i] (-10000 ≤ tY[i] ≤ 10000)

第四行为办公室坐标gx,gy(-10000 ≤ gx,gy ≤ 10000),以空格分隔

第五行为走路时间walkTime(1 ≤ walkTime ≤ 1000)和taxiTime(1 ≤ taxiTime ≤ 1000),以空格分隔

输出描述:

输出一个整数表示，小易最快能赶到办公室的时间

输入例子:

2
-2 -2
0 -2
-4 -2
15 3

输出例子:

42

我的答案：
import java.awt.Point;
import java.util.Scanner;
 
 
public class Main {
	
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
    	int n = in.nextInt();
    	Point taxi[] = new Point[n];
    	for(int i=0;i<n;i++){
    		taxi[i] = new Point();
    	}
    	for(int i=0;i<n;i++){
    		int x = in.nextInt();
    		taxi[i].x = x; 
    	}
    	for(int i=0;i<n;i++){
    		int y = in.nextInt();
    		taxi[i].y = y;
    	}
    	int x = in.nextInt();
    	int y = in.nextInt();
    	Point end = new Point(x, y);
    	Point start = new Point(0,0);
    	int WalkSpeed = in.nextInt();
    	int CarSpeed = in.nextInt();
    	int walk_distance = Distance(start,end);
    	int walk_time = walk_distance*WalkSpeed;
    	int car_time = Integer.MAX_VALUE;
    	for(int i=0;i<n;i++){
    		int time = Distance(start,taxi[i])*WalkSpeed + Distance(end,taxi[i])*CarSpeed;
    		if(time<car_time)
    			car_time = time;
    	}
    	System.out.println(Math.min(walk_time, car_time));
    }
 
	private static int Distance(Point start, Point end) {
		return Math.abs(start.x-end.x)+Math.abs(start.y-end.y);
	}
    
}