poj3624Charm Bracelet

最新推荐文章于 2022-04-25 19:27:35 发布

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最新推荐文章于 2022-04-25 19:27:35 发布

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分类专栏： 01背包动态规划文章标签： poj 01背包

本文链接：https://blog.youkuaiyun.com/z8110/article/details/48557689

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本文介绍了一道关于选择最佳组合的魅力手链问题。在给定重量限制和一系列魅力物件的重量与吸引力评分的情况下，需要找到最大总吸引力评分的组合。通过使用类似于01背包问题的动态规划方法解决此问题。

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Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 27837		Accepted: 12534

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

Source

USACO 2007 December Silver

其实题不是看的懂，看输出有点像01背包，然后就a了。

附代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
long ans[1100000],weight[1100000],weath[1100000],i,j,k,l,m,n;//数组要开的大 
int main()
{
	
	while(scanf("%ld%ld",&m,&n)!=EOF)
	{
		
	memset(ans,0,sizeof(ans));
	for(i=1;i<=m;i++)
	scanf("%ld%ld",&weight[i],&weath[i]);
	
	
	for(i=1;i<=m;i++)
	for(j=n;j>=weight[i];j--)
	ans[j]=max(ans[j],ans[j-weight[i]]+weath[i]);
	printf("%ld\n",ans[n]);
	}
}