Charm Bracelet（01背包）

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本文介绍了一种经典的计算机科学问题——01背包问题，并通过一个具体的示例详细讲解了如何解决此类问题。文章中提供了完整的C语言代码实现，旨在帮助读者理解如何在给定重量限制下选择物品以达到价值最大化。

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18028		Accepted: 8237

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：

给出N（1到3402）颗钻石，M（1到12880）总容量。再给出N颗每颗钻石的w（1到12880）容量和v价值。求在不超过总容量M的情况下往包放钻石的最大价值。

思路：

01背包问题。没有特殊条件（恰好装满），只要求求最大值故初始化为0。

AC：

#include<stdio.h>
#include<string.h>
int w[4000],d[4000];
int dp[15000];

int main()
{
	int n,m;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
	memset(dp,0,sizeof(dp));
	for(int i=1;i<=n;i++)
		scanf("%d%d",&w[i],&d[i]);
	for(int i=1;i<=n;i++)
		for(int j=m;j>=w[i];j--)
			dp[j]=(dp[j]>dp[j-w[i]]+d[i]?dp[j]:dp[j-w[i]]+d[i]);
	printf("%d\n",dp[m]);
	}
	return 0;
}