Charm Bracelet-POJ3624(01背包)

http://poj.org/problem?id=3624

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 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

这个是简单的01背包   从我开始接触到背包到现在我还是不懂背包的原理 开始背包的旅程正常的背包是i     1....nj     m....0dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+v[i]);但是这个范围较大如果用二维的会超内存可以转化成一维的dp[j]=max(dp[j],dp[j-w[i]]+v[i]);

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<iostream>

using namespace std;

#define N 3500
int dp[13000];

int main()
{
    int n,m,w[N],v[N];
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%d %d",&w[i],&v[i]);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            for(int j=m;j>=0;j--)
            {
                if(j>=w[i])
                   dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
            }
        }
        printf("%d\n",dp[m]);
    }
    return 0;
}