L - Charm Bracelet POJ - 3624

最新推荐文章于 2020-05-18 22:19:45 发布

Neng_Miao

最新推荐文章于 2020-05-18 22:19:45 发布

阅读量246

点赞数

CC 4.0 BY-SA版权

分类专栏： 2018寒假第二周

本文链接：https://blog.youkuaiyun.com/Seven_deadly_sins/article/details/79226814

2018寒假第二周专栏收录该内容

4 篇文章

订阅专栏

本文介绍了一个经典的01背包问题，通过示例讲解了如何利用动态规划算法来解决该问题，包括输入输出样例及核心代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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n种物品的体积和价值，在V体积下所能装的最大价值是多少，01背包简单问题。

#include<stdio.h>
#include<string.h>
int Max(int x,int y)
{
	if(x>y)
		return x;
	return y;
}
int main()
{
	int i,j,k,m,n,v;
	int f[5000],w[5000];
	int c[15000]; 
	while(scanf("%d%d",&m,&v)!=EOF)
	{
		memset(c,0,sizeof(c));
		memset(f,0,sizeof(f));
		memset(w,0,sizeof(w)); 
		for(i=1;i<=m;i++)
			scanf("%d%d",&f[i],&w[i]);
		
		for(i=1;i<=m;i++)
		{
			for(j=v;j>=f[i];j--)
			{
				c[j]=Max(c[j],c[j-f[i]]+w[i]);//动态方程，对于每个体积状态下判断当前体积上能否到最大价值
			}
		}
		printf("%d\n",c[v]);
	}
	return 0;
}