POJ 3624（简单01背包）

最新推荐文章于 2022-03-11 01:34:41 发布

Sun_AC

最新推荐文章于 2022-03-11 01:34:41 发布

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分类专栏： dp POJ 文章标签： poj 01背包

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本文介绍了一个经典的背包问题案例，通过动态规划的方法解决如何在给定的重量限制下选择最有价值的物品组合。提供了完整的代码实现。

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题目连接：http://poj.org/problem?id=3624

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23411		Accepted: 10551

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

AC代码：

#include <cstdio>
#define max(a,b) a>b?a:b;
int n,m,w[3500],p[3500],f[13000],i,j;
int main(){
	scanf("%d%d",&n,&m);
	for(i=0;i<n;i++) scanf("%d%d",&w[i],&p[i]);
	for(i=0;i<n;i++)
		for(j=m;j>=w[i];j--)
			f[j]=max(f[j],f[j-w[i]]+p[i]);
	printf("%d\n",f[m]);
	return 0;
}