POJ 3624 Charm Bracelet（01背包）

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#poj #01背包 #动态规划

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本文介绍了一道经典的背包问题——最优魅力手链问题。题目要求在给定重量限制下选择若干个魅力饰品以达到最高的魅力值总和。文章通过一个具体的示例详细解释了如何使用动态规划解决这个问题，并给出了完整的C++实现代码。

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题目链接：http://poj.org/problem?id=3624

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 44997		Accepted: 19254

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

Sample Output

Source

USACO 2007 December Silver

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#define maxn 13500
using namespace std;

int i, j, n, m;
int w[maxn], v[maxn], dp[maxn];

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {

        for(i=1;i<=n;i++)
            cin>>w[i]>>v[i];

        for(i=1;i<=n;i++)
            for(j=m;j>=w[i];j--)
                dp[j]=max(dp[j],dp[j-w[i]]+v[i]);

        cout<<dp[m]<<endl;
    }
    return 0;
}