Problem:
0-1 Knapsack Problem

You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.

You want to find a subset of items to put such that:

The total value of the items is as large as possible.

The items have combined weight at most W, that is capacity of the knapsack.

Find the maximum total value of items in the knapsack.

Input

N W
v1 w1
v2 w2
:
vN wN

The first line consists of the integers N and W. In the following lines, the value and weight of the i-th item are given.

Output

Print the maximum total values of the items in a line.

Constraints

1 ≤ N ≤ 100

1 ≤ vi ≤ 1000

1 ≤ wi ≤ 1000

1 ≤ W ≤ 10000

Sample Input 1

4 5
4 2
5 2
2 1
8 3

Sample Output 1

13

Sample Input 2

2 20
5 9
4 10

Sample Output 2

9

Solution in Scala (judge verdict: Accepted)

import scala.io.Source

object Main {
  def main(args: Array[String]): Unit = {
    val lines = Source.stdin.getLines.toStream
    val Array(numLines, weightMax) = lines.head.split(" ").map(_.toInt)
    val valueWeight = lines.drop(1).take(numLines).map{ v =>
      val Array(value, weight) = v.split(" ").map(_.toInt)
      (value, weight)
    }.toArray
    val memo = (new Array[Int]((numLines+1) * (weightMax+1))).map(_ => -1)
    println(dp(numLines - 1, weightMax, valueWeight, memo, numLines, weightMax))
    // println(memo.toList)
  }

  def dp(i: Int, j: Int, arr: Array[(Int, Int)], memo: Array[Int], width: Int, height: Int): Int = {
    if (i < 0 || j == 0) {
      0
    } else if (memo(width*j + i) != -1) {
      memo(width*j + i)
    } else {
      if (j >= arr(i)._2) {
        memo(width*j + i) =
          math.max(
            dp(i - 1, j, arr, memo, width, height),
            dp(i - 1, j - arr(i)._2, arr, memo, width, height) + arr(i)._1
          )
      } else {
        memo(width*j + i) = dp(i - 1, j, arr, memo, width, height)
      }
      if (memo(width*j + i) == -1) memo(width*j + i) = 0
      memo(width*j + i)
    }
  }
}