POJ 2970 The lazy programmer(贪心+单调优先队列)

A new web-design studio, called SMART (Simply Masters of ART), employs two people. The first one is a web-designer and an executive director at the same time. The second one is a programmer. The director is so a nimble guy that the studio has already got N contracts for web site development. Each contract has a deadline d_i.

It is known that the programmer is lazy. Usually he does not work as fast as he could. Therefore, under normal conditions the programmer needs bi of time to perform the contract number i. Fortunately, the guy is very greedy for money. If the director pays him x_i dollars extra, he needs only (b_i − a_i x_i) of time to do his job. But this extra payment does not influent other contract. It means that each contract should be paid separately to be done faster. The programmer is so greedy that he can do his job almost instantly if the extra payment is (b_i ⁄ a_i) dollars for the contract number i.

The director has a difficult problem to solve. He needs to organize programmer’s job and, may be, assign extra payments for some of the contracts so that all contracts are performed in time. Obviously he wishes to minimize the sum of extra payments. Help the director!

Input

The first line of the input contains the number of contracts N (1 ≤ N ≤ 100 000, integer). Each of the next N lines describes one contract and contains integer numbers a_i, b_i, d_i (1 ≤ a_i, b_i ≤ 10 000; 1 ≤ d_i ≤ 1 000 000 000) separated by spaces.

Output

The output needs to contain a single real number S in the only line of file. S is the minimum sum of money which the director needs to pay extra so that the programmer could perform all contracts in time. The number must have two digits after the decimal point.

Sample Input

2
20 50 100
10 100 50

Sample Output

5.00

 

题意：  有n的任务，每个任务有自己是时长bi，截止时期di，和1单位钱可以减少的时间ai。 问为使所有任务都可以在自己截止前完成，所需要的最少的钱。

思路：  显然是贪心题，但是单纯以时间排序或者以ai排序都是不合理的，可以反证。 

             可以以时间排序，然后每次先加到队列里，如果超过截止日期，再在队列里找代价（ai）最小的。