POJ 3624 Charm Bracelet

最新推荐文章于 2019-08-06 15:11:09 发布

转载最新推荐文章于 2019-08-06 15:11:09 发布 · 69 阅读

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原文链接：http://www.cnblogs.com/Loi-Vampire/p/6017049.html

本文介绍了一个简单的背包问题实例，通过一维数组实现最优解的计算。该问题要求在给定重量限制下，从一系列具有不同权重和吸引力的装饰品中选择最佳组合，以最大化总吸引力。文章提供了一段C++代码示例，展示了如何使用动态规划解决此类问题。

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 Wi (1 ≤ Wi ≤ 400), a ‘desirability’ factor Di (1 ≤ Di ≤ 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: 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

4 6
1 4
2 6
3 12
2 7

Sample Output

23

Source

USACO 2007 December Silver

简单背包，但是…数据范围二维数组开不起，所以要用一维数组。

#include <iostream>
#include <cstdio>
using namespace std;

struct DC
{
    int w, d;
}p[3450];
int dp[13300];

int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++)
        scanf("%d%d", &p[i].w, &p[i].d);
    for(int i = 1; i <= n; i++)
        for(int j = m; j >= p[i].w; j--)
            dp[j] = max(dp[j - p[i].w] + p[i].d, dp[j]);
    cout<<dp[m];
    return 0;
}

转载于:https://www.cnblogs.com/Loi-Vampire/p/6017049.html