POJ3624(01背包:滚动 实现)

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30417		Accepted: 13576

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

#include"cstdio"
#include"cstring"
#include"algorithm"
using namespace std;
const int MAXN=3405;
int dp[13000];
int n,W;
int v[MAXN],w[MAXN];
int main()
{
    while(scanf("%d%d",&n,&W)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++)    scanf("%d%d",&w[i],&v[i]);
        
        for(int i=0;i<n;i++)
        {
            for(int j=W;j>=w[i];j--)    dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
        }
        printf("%d
",dp[W]);
    }
    
    return 0;
}