01背包问题：poj 3624 Charm Bracelet-优快云博客

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，以达到最高的价值总和。文章详细展示了如何初始化数组、进行状态转移，并最终输出最优解。

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Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 25953		Accepted: 11672

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

刚开始的时候数组开到5000，结果RE了，改大数组之后就AC了。

#include<iostream>
#include<cstring>
using namespace std;
const int MAX=20000;
int dp[MAX];
int main()
{
	int N,M;
	while(cin>>N>>M)
	{
	
		int w[MAX];
		int v[MAX];
		memset(dp,0,sizeof(dp));
		for(int i =1 ;i <= N; i ++ )
		    cin>>w[i]>>v[i];
                for(int i = 1; i <= N;i ++){
        	     for(int j =M ;j >= w[i]; j--){
        		dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
        	}
        }
        cout<<dp[M]<<endl;
	}
	return 0;
}