poj--3624--Charm Bracelet（动态规划水题）

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最新推荐文章于 2022-03-25 21:17:10 发布

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分类专栏：动态规划

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本文介绍了一种基于动态规划解决的魅力手链问题。问题描述了如何在给定重量限制下选择最佳组合的装饰物以最大化魅力值。文章提供了完整的代码实现，并通过样例输入输出展示了算法的有效性。

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POJ - 3624

Charm Bracelet

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

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

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[100100];
struct node
{
	int w,val;
}edge[100100];
int main()
{
	int m,n;
	while(scanf("%d%d",&m,&n)!=EOF)
	{
		memset(dp,0,sizeof(dp));
		for(int i=0;i<m;i++)
		scanf("%d%d",&edge[i].w,&edge[i].val);
		for(int i=0;i<m;i++)
		for(int j=n;j>=edge[i].w;j--)
		dp[j]=max(dp[j],dp[j-edge[i].w]+edge[i].val);
		printf("%d\n",dp[n]);
	}
	return 0;
}

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