poj-3624-Charm Bracelet【经典01背包】

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Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 34487		Accepted: 15275

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

大意：基础背包问题，大概意思是，承重 m 的背包，有 n 个物品，第 i 个物品重 w [ i ]，价值 val [ i ]，求背包能装入物品的最大价值

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n,m;
int w[5000],val[5000];
int dp[13000];
int main()
{
  while(~scanf("%d%d",&n,&m))
  {
    for(int i=1;i<=n;i++)
    {
      scanf("%d%d",&w[i],&val[i]);
    }
    memset(dp,0,sizeof(dp));
    for(int i=1;i<=n;i++)
    {
      for(int j=m;j>=w[i];j--)
      {
        dp[j]=max(dp[j],dp[j-w[i]]+val[i]);
      }
    }
    printf("%d\n",dp[m]);
  }
  return 0;
 }