POJ 3624 还是01背包

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本文介绍了一个经典的背包问题求解案例，通过动态规划算法实现，在给定的重量限制下选取一系列物品，使得总价值最大化。文章提供了完整的代码示例，帮助读者理解如何在限定条件下寻找最优解。

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

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 weightW_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

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
int f[20005],w[20005],v[20005];
int main()
{
    int n,V;
    while(cin>>n>>V)
    {
        memset(f,0,sizeof(f));// 先清零是很必要的
        memset(w,0,sizeof(w));
        memset(v,0,sizeof(v));
        for(int i=1; i<=n; i++)
            cin >> w[i]>>v[i];
        for(int i=1; i<=n; i++)
            for(int j=V; j>=w[i]; j--)
                f[j]=max(f[j],f[j-w[i]]+v[i]);
        cout << f[V]<< endl;
    }
    return 0;
}