论代码质量和效率,这个题自己写了代码实现了和网上的大神的代码对比了下,代码技巧还是有待提高(要短少精炼的好代码啊!)。。。。,后面分别放上本人和大神的代码:
终于到周末啦!小易走在市区的街道上准备找朋友聚会,突然服务器发来警报,小易需要立即回公司修复这个紧急bug。假设市区是一个无限大的区域,每条街道假设坐标是(X,Y),小易当前在(0,0)街道,办公室在(gx,gy)街道上。小易周围有多个出租车打车点,小易赶去办公室有两种选择,一种就是走路去公司,另外一种就是走到一个出租车打车点,然后从打车点的位置坐出租车去公司。每次移动到相邻的街道(横向或者纵向)走路将会花费walkTime时间,打车将花费taxiTime时间。小易需要尽快赶到公司去,现在小易想知道他最快需要花费多少时间去公司。
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
我的代码:
def walk_time(gx, gy, walkTime):
return (abs(gx) +abs(gy)) * walkTime
def taxi_time(gx, gy, lt1, lt2, walkTime, taxiTime):
time_spend = []
while len(lt1):
t_x = lt1.pop()
t_y = lt2.pop()
time1 = (abs(t_x) + abs(t_y)) * walkTime
time2 = (abs(gx - t_x) + abs(gy - t_y)) * taxiTime
time_spend.append(time1 + time2)
spend = min(time_spend)
return spend
if __name__ == "__main__":
n = int(raw_input("input n:"))
ls_x = [int(i)for i in raw_input("input x coordinate list:").strip().split()]
ls_y = [int(i) for i in raw_input("input y coordinate list:").strip().split()]
g = [int(i) for i in raw_input("input gx and gy:").strip().split()]
gx = g[0]
gy = g[1]
T = [int(i) for i in raw_input("input walkTime and taxiTime:").strip().split()]
walkTime = T[0]
taxiTime = T[1]
print(min(taxi_time(gx, gy, ls_x, ls_y, walkTime, taxiTime), walk_time(gx, gy, walkTime)))
复现大神的实现:
import sys
while True:
line = sys.stdin.readline().strip("\n")
if line.strip() == "":
break
num = int(line)
x = [int(val) for val in sys.stdin.readline().split(" ")]
y = [int(val) for val in sys.stdin.readline().split(" ")]
locate = [int(val) for val in sys.stdin.readline().split(" ")]
speed = [int(val) for val in sys.stdin.readline().split(" ")]
time = [speed[0] * (abs(locate[0]) + abs(locate[1]))]
for i in range(num):
time.append(speed[0] * (abs(x[i]) + abs(y[i])) + speed[1] * (abs(locate[0] - x[i]) + abs(locate[1] - y[i])))
print min(time)