【POJ3264】 Charm Bracelet

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 theN (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weightW_i (1 ≤ W_i ≤ 400), a 'desirability' factorD_i (1 ≤ D_i ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more thanM (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 61 42 63 122 7

Sample Output

23

#include <cstring>

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <map>
#include <queue>
using namespace std;

int N,M;
int w[4000];
int v[4000];
int dp[4000][400];
int main()
{
    while(scanf("%d%d",&N,&M) != EOF)
    {
        memset(dp,0,sizeof(dp));
        memset(w,0,sizeof(w));
        memset(v,0,sizeof(v));
        for(int i=1;i<=N;i++)
        {
            scanf("%d%d",&w[i],&v[i]);
        }

        for(int i=1;i<=N;i++)
        {
            for(int j=M;j>=0;j--)
            {
                if(j >= w[i])
                dp[i][j] = max(dp[i-1][j],dp[i-1][j-w[i]] + v[i]);
            }
        }
        for(int i=0;i<=N;i++)
        {
            for(int j=0;j<=M;j++)
        {
            cout << dp[i][j] << " " ;
        }
        }
        printf("%d\n",dp[N][M]);
    }
    return 0;
}