Problem:
0-1 Knapsack Problem II

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 N 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 ≤ 100

1 ≤ wi ≤ 10,000,000

1 ≤ W ≤ 1,000,000,000

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 java.util.Scanner

object Main extends App { 
  type Item = (Int, Int)
  class DP(num:Int, wgt:Int, items:Array[Item]) {
    val inf = Int.MaxValue >> 1
    val sumOfValue = items.foldLeft(0)((sum:Int, item:Item) => { val (v,w)=item; v+sum })
    val memo = Array.fill(num+1)(Array.fill(sumOfValue+1)(inf))
    memo.foreach((a)=>a(0)=0)
    def solver():Int = {
      var maxValue = 0 
      for (i <- 1 to num; j <- 1 to sumOfValue) {
        val (v, w) = items(i-1)
        val w0 = memo(i-1)(j)
        val w1 = if (j < v) w else memo(i-1)(j-v) + w
        val wmin = w0 min w1
        if (wmin <= wgt) maxValue = maxValue max j
        memo(i)(j)= wmin
      }
      maxValue
    }
  }

  val sc = new Scanner(System.in)
  val N = sc.nextInt
  val W = sc.nextInt
  val items = (for (i <- 1 to N) yield (sc.nextInt, sc.nextInt)).toArray

  val dp = new DP(N, W, items)
  
  println(dp.solver())

}