Problem:
Score : 400 points

Problem Statement

There are N pieces of sushi. Each piece has two parameters: "kind of topping" t_i and "deliciousness" d_i.
You are choosing K among these N pieces to eat.
Your "satisfaction" here will be calculated as follows:

The satisfaction is the sum of the "base total deliciousness" and the "variety bonus".

The base total deliciousness is the sum of the deliciousness of the pieces you eat.

The variety bonus is x*x, where x is the number of different kinds of toppings of the pieces you eat.

You want to have as much satisfaction as possible.
Find this maximum satisfaction.

Constraints

1 \leq K \leq N \leq 10^5

1 \leq t_i \leq N

1 \leq d_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
t_1 d_1
t_2 d_2
.
.
.
t_N d_N

Output

Print the maximum satisfaction that you can obtain.

Sample Input 1

5 3
1 9
1 7
2 6
2 5
3 1

Sample Output 1

26

If you eat Sushi 1,2 and 3:

The base total deliciousness is 9+7+6=22.

The variety bonus is 2*2=4.

Thus, your satisfaction will be 26, which is optimal.

Sample Input 2

7 4
1 1
2 1
3 1
4 6
4 5
4 5
4 5

Sample Output 2

25

It is optimal to eat Sushi 1,2,3 and 4.

Sample Input 3

6 5
5 1000000000
2 990000000
3 980000000
6 970000000
6 960000000
4 950000000

Sample Output 3

4900000016

Note that the output may not fit into a 32-bit integer type.

Solution in Scala (2.11.7) (judge verdict: Wrong Answer)


object Main {

  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)

    val N, K = sc.nextInt()

    case class Sushi(t: Int, d: Long)

    val _Sushis = new Array[Sushi](N)
    (0 until N) foreach { i =>
      val t = sc.nextInt()
      val d = sc.nextLong()
      _Sushis(i) = Sushi(t, d)
    }
    val Sushis = _Sushis.sortBy(_.d)(Ordering.Long.reverse)

    // f(a)とそれを構成するS_aがわかっている時に
    // f(a+1)とそれを構成するS_a+1をを求めることを考えます(x <= a <= N-1)
    // S_a+1はS_aに以下の操作を加えることで構成することができます
    // 1. 集合S_aから、取り除いても「ネタ」の種類数が減らない寿司の中で最も「おいしさ」の小さいものを取り除く
    //    この時点でネタの種類数が減らないものがない場合には, それまでのものから選ぶ
    // 2. 集合S_aに含まれない寿司の中で、集合S_aにない「ネタ」を持つ、最も「おいしさ」の大きいものをS_aに加える
    // この操作を繰り返すことにより、f(i) (x < i <= N)を再帰的に求めることができます。

    val (initialSushis, restSushis) =
      Sushis.sorted(
        Ordering.by[Sushi, (Int, Long)](s => (s.t, s.d))(
          Ordering.Tuple2(Ordering.Int, Ordering.Long.reverse)))
        .splitAt(K)

    val sushiset = initialSushis.map(_.t).toSet
    var p = sushiset.size.toLong
    val initialV = initialSushis.foldLeft(0L)(_ + _.d) + p * p

    val removableSushis = scala.collection.mutable.PriorityQueue[Sushi]()(Ordering.by[Sushi, Long](_.d).reverse)
    val appendableSushis = scala.collection.mutable.PriorityQueue[Sushi]()(Ordering.by[Sushi, Long](_.d))

    {
      val N = initialSushis.length
      var bef = -1
      var i = 0
      while (i < N) {
        val sushi = initialSushis(i)
        if (sushi.t == bef) {
          removableSushis.enqueue(sushi)
        }
        bef = sushi.t
        i += 1
      }
    }

    {
      val N = restSushis.length
      var bef = -1
      var i = 0
      while (i < N) {
        val sushi = restSushis(i)
        if (!sushiset.contains(sushi.t)) {
          if (sushi.t != bef) {
            appendableSushis.enqueue(sushi)
          }
          bef = sushi.t
        }
        i += 1
      }
    }

    def solve(bef: Long, max: Long, p: Long): Long =
      if (p < K && removableSushis.nonEmpty && appendableSushis.nonEmpty) {
        val rs = removableSushis.dequeue()
        val as = appendableSushis.dequeue()
        val current = bef - rs.d + as.d + 2 * p + 1
        solve(current, max max current, p + 1)
      } else max

    println(solve(initialV, initialV, p))
  }

}