poj3624 Charm Bracelet

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1783		Accepted: 780

Description

Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight W_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).

Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: W_i and D_i

Output

* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

Sample Input

4 6
1 4
2 6
3 12
2 7

Sample Output

23

Source

USACO 2007 December Silver

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//5037530 11410 3624 Accepted 284K 250MS C++ 693B 2009-04-24 13:59:01
#include <iostream>
#define MAX 12882
using namespace std;
struct node
{
 int w;
 int v;
}data[MAX];
int dp[MAX],n,m;
void Init()
{
 int i;
 for(i=1;i<=n;i++)
  scanf("%d%d",&data[i].w,&data[i].v);
}
int GetMax(int a,int b)
{
 if(a>b)
  return a;
 else
  return b;
}
void Knapsack()
{
 int i,j;
 for(i=0;i<=m;i++)
 {
  if(i>=data[n].w)
   dp[i]=data[n].v;
  else
   dp[i]=0;
 }
 for(i=n-1;i>=1;i--)
  for(j=m;j>=1;j--)
  {
   if(j>=data[i].w)
   {
    dp[j]=GetMax(dp[j],dp[j-data[i].w]+data[i].v);
   }
   else
    break;
  }
}
int main()
{
 while(scanf("%d%d",&n,&m)!=EOF)
 {
  Init();
  Knapsack();
  printf("%d\n",dp[m]);
 }
 return 0;
}