poj3624Charm Bracelet

最新推荐文章于 2021-01-22 22:57:48 发布

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最新推荐文章于 2021-01-22 22:57:48 发布

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文章标签：数据结构与算法

本文介绍了通过动态规划解决01背包问题，旨在找到给定重量限制和物品权重与价值列表下，最大可能的价值总和。

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23025		Accepted: 10358

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

题意：01背包

重点：01背包、动态规划

难点：动态方程

#include<cstdio>
#include<cstring>
int main()
{
	int n,s,i,j,v[40000],w[40000],dp[40000];
	while(scanf("%d%d",&n,&s)==2)
	{
		memset(v,0,sizeof(v));
		memset(w,0,sizeof(w));
		memset(dp,0,sizeof(dp));

		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&w[i],&v[i]);
		}
		for(i=1;i<=n;i++)
			for(j=s;j>=w[i];j--)
			{
				if(dp[j]<dp[j-w[i]]+v[i])
					dp[j]=dp[j-w[i]] + v[i] ;
			} 
		printf("%d\n",dp[s]);
	}return 0;
}