POJ 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;
}