POJ—— 3624 Charm Bracelet（01背包）

最新推荐文章于 2021-01-22 22:57:48 发布

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最新推荐文章于 2021-01-22 22:57:48 发布

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分类专栏： ACM解题报告文章标签：背包问题 POJ

本文链接：https://blog.youkuaiyun.com/hunt_er/article/details/79183510

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本文介绍了一个经典的01背包问题实例，通过一个具体的示例详细解释了如何利用动态规划求解最大魅力值的问题。文章提供了完整的C语言实现代码，并通过样例输入输出展示了算法的有效性和正确性。

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题目链接：http://poj.org/problem?id=3624

题目：

Charm Bracelet

Problem 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 theN (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weightWi (1 ≤ Wi ≤ 400), a 'desirability' factorDi (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more thanM (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:Wi and Di

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

题目描述：

01背包问题。

代码：

#include<stdio.h> 
#include<string.h>
int f[15000];
int e[3500];
int w[3500];
int a,b,c,i,j;
int main()
{
	int t;
	while(scanf("%d%d",&a,&b)!=EOF){
		for(i = 0;i < a;i ++)
			scanf("%d%d",&e[i],&w[i]);
		memset(f,0,sizeof(f));
		for(i = 0;i < a;i ++){
			for(j = b;j >= e[i];j--){
				f[j] = f[j] > f[j - e[i]] + w[i] ? f[j] : f[j - e[i]] + w[i];
			}
		}
		printf("%d\n",f[b]);
	}
	return 0;
}