题目有10辆各种类型汽车到加油站加油,设加油站加满第i(i=1,2,…,10)辆车的油所需要的时间为T_i分,这些T_i各不相同,如果每次只对一辆汽车加油,应该如何安排加油车辆的顺序,使花费的总时间最少?(应计算各车加油时间,等待时间)最少时间等于多少? 解先考虑简单情形:假定只有3辆汽车排队加油,T_i=2分,T_i=3分,T_i=4分.不难猜测和证明,加油时间少的汽车先加油时,总时间最少.这是因为后面的汽车等待的时间最少.这时排队加油的总时间为2+(2+3)+(2+3+4)(分). There are 10 vehicles of various types to the gas station to refuel, and the time required for the filling station to fill up the oil of the i-th (i=1, 2, ..., 10) vehicles is T_i points. These T_i are different, if How to arrange the order of refueling vehicles at a time, and how to arrange the refueling vehicles so that the total time spent is at least? (The refueling time and waiting time should be calculated.) What is the minimum time equal? ​​Consider the simple case: Suppose there are only 3 vehicles. Cars line up for fuel, T_i = 2 points, T_i = 3 points, and T_i = 4 points. It’s not hard to guess and prove that when the car with less refuel time is refueled, the total time is the least. This is because the back car has the least waiting time. The total time to queue for refueling is 2+(2+3)+(2+3+4)(minutes).