Codeforces Round #433 (Div. 2)
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题意:
给你n个飞机,k。按出发顺序1-n给你每个飞机拖延1分钟所要给的分数。让你重新排列飞机的飞行表。
前1-k分钟不能起飞。求一个出发顺序,让总分数最少。
POINT:
贪心安排,先安排分数最多的。
安排的时候用链表来连接分钟。不然一个一个判断超时。
C. Planningtime limit per test1 secondmemory limit per test512 megabytesinputstandard inputoutputstandard outputHelen works in Metropolis airport. She is responsible for creating a departure schedule. There are n flights that must depart today, the i-th of them is planned to depart at the i-th minute of the day.Metropolis airport is the main transport hub of Metropolia, so it is difficult to keep the schedule intact. This is exactly the case today: because of technical issues, no flights were able to depart during the first k minutes of the day, so now the new departure schedule must be created.All n scheduled flights must now depart at different minutes between (k + 1)-th and (k + n)-th, inclusive. However, it's not mandatory for the flights to depart in the same order they were initially scheduled to do so — their order in the new schedule can be different. There is only one restriction: no flight is allowed to depart earlier than it was supposed to depart in the initial schedule.Helen knows that each minute of delay of the i-th flight costs airport ci burles. Help her find the order for flights to depart in the new schedule that minimizes the total cost for the airport.InputThe first line contains two integers n and k (1 ≤ k ≤ n ≤ 300 000), here n is the number of flights, and k is the number of minutes in the beginning of the day that the flights did not depart.The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 107), here ci is the cost of delaying the i-th flight for one minute.OutputThe first line must contain the minimum possible total cost of delaying the flights.The second line must contain n different integers t1, t2, ..., tn (k + 1 ≤ ti ≤ k + n), here ti is the minute when the i-th flight must depart. If there are several optimal schedules, print any of them.Exampleinput5 24 2 1 10 2output203 6 7 4 5 NoteLet us consider sample test. If Helen just moves all flights 2 minutes later preserving the order, the total cost of delaying the flights would be (3 - 1)·4 + (4 - 2)·2 + (5 - 3)·1 + (6 - 4)·10 + (7 - 5)·2 = 38 burles.However, the better schedule is shown in the sample answer, its cost is (3 - 1)·4 + (6 - 2)·2 + (7 - 3)·1 + (4 - 4)·10 + (5 - 5)·2 = 20 burles.
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- Codeforces Round #433 (Div. 2)
- Codeforces Round #433 (Div. 2)
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