本文主要是介绍poj 3624Charm Bracelet(简单01背包)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
这是一道水的不能再水的题了，不过，不用空间优化的话，会超内存的……
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10038 | Accepted: 4511 |
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 Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 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: Wi and Di
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
#include<iostream>
using namespace std;
int a[12881]={0};
int main()
{int n, m;int w[3403], d[3403];cin>>n>>m;int i, j;for(i=1; i<=n; i++)cin>>w[i]>>d[i];for(i=1; i<=n; i++)for(j=m; j>=w[i]; j--){if( a[j]<a[j-w[i]]+d[i])a[j]=a[j-w[i]]+d[i];}cout<<a[m]<<endl;
}
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