HDU 3624 Charm Bracelet （01背包）

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13977		Accepted: 6381

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

Source

USACO 2007 December Silver

 

 

 

简单的01背包，

开始练习背包

 

 

/*

简单的01背包

*/

#include<stdio.h>

#include<iostream>

#include<string.h>

#include<algorithm>

using namespace std;



const int MAXN=3500;

int dp[13000];

int value[MAXN];//每袋的价格

int weight[MAXN];//每袋的重量



int nValue,nKind;



//0-1背包，代价为cost,获得的价值为weight

void ZeroOnePack(int cost,int weight)

{

    for(int i=nValue;i>=cost;i--)

      dp[i]=max(dp[i],dp[i-cost]+weight);

}



int main()

{

    while(scanf("%d%d",&nKind,&nValue)!=EOF)

    {

        for(int i=0;i<nKind;i++)scanf("%d%d",&value[i],&weight[i]);

        memset(dp,0,sizeof(dp));

        for(int i=0;i<nKind;i++)

          ZeroOnePack(value[i],weight[i]);

        printf("%d\n",dp[nValue]);

    }

    return 0;

}