poj 3624 (背包入门)

入门题，最近又一次想不起来背包的dp方法，无奈又得找道水题找找感觉...

Charm Bracelet

Time Limit:1000MS		Memory Limit:65536K
Total Submissions:15063		Accepted:6880

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 charmiin 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:NandM
* Lines 2..N+1: Linei+1 describes charmiwith two space-separated integers:W_iandD_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

Sample Output

Source

USACO 2007 December Silver

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
#define N 13000
int dp[N];

int main()
{
    int n,m;
    memset(dp,0,sizeof(dp));
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++)
    {
        int tmp,w;
        scanf("%d%d",&w,&tmp);
        for(int j=m;j>=w;j--)
        {
            dp[j]=max(dp[j],dp[j-w]+tmp);
        }
    }
    printf("%d\n",dp[m]);
    return 0;
}

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