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最新推荐文章于 2025-09-09 08:37:13 发布

转载最新推荐文章于 2025-09-09 08:37:13 发布 · 118 阅读

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原文链接：http://www.cnblogs.com/jlyg/p/6644947.html

文章标签：

#java

本文针对一个具体的购物问题，通过分析不同商品与箱子的关系及其价值，采用动态规划算法解决如何在有限预算内获得最大价值的问题。文章详细介绍了算法的实现过程，并给出了一段使用滚动数组优化的AC代码。

这道题真正困扰了笔者3,4天，冥思苦想几日无果之后，只能去找大牛的解法。结合网上的大牛解法与自己的理解，笔者终于解决了这个坑了，在此小庆幸一下。

原题如下：

Consumer
Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/65536 K (Java/Others)
Total Submission(s): 2154    Accepted Submission(s): 1157


Problem Description
FJ is going to do some shopping, and before that, he needs some boxes to carry the different kinds of stuff he is going to buy. Each box is assigned to carry some specific kinds of stuff (that is to say, if he is going to buy one of these stuff, he has to buy the box beforehand). Each kind of stuff has its own value. Now FJ only has an amount of W dollars for shopping, he intends to get the highest value with the money.

 

Input
The first line will contain two integers, n (the number of boxes 1 <= n <= 50), w (the amount of money FJ has, 1 <= w <= 100000) Then n lines follow. Each line contains the following number pi (the price of the ith box 1<=pi<=1000), mi (1<=mi<=10 the number goods ith box can carry), and mi pairs of numbers, the price cj (1<=cj<=100), the value vj(1<=vj<=1000000)

 

Output
For each test case, output the maximum value FJ can get

 

Sample Input
3 800
300 2 30 50 25 80
600 1 50 130
400 3 40 70 30 40 35 60
 

Sample Output
210

View Code

AC代码如下：其中dp使用了滚动数组。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int dp[2][100010];
int main()
{
    int n,w;
    while(~scanf("%d%d",&n,&w))
    {
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;++i)
        {
            int pi,mi;
            scanf("%d%d",&pi,&mi);
            //这里做了个假设，假设dp[1][j]组必须加本轮的箱子
            //得到结果后和上一轮的结果比
            for(int j=0;j<=w-pi;++j)
                dp[1][j+pi] = dp[0][j];
            for(int j=0;j<mi;++j)
            {
                int cj,wj;
                scanf("%d%d",&cj,&wj);
                //当v-cj小于pi，表示连箱子都买不起
                for(int v=w;v-cj>=pi;--v)
                {
                    //01背包
                    dp[1][v] = max(dp[1][v],dp[1][v-cj]+wj);
                }
            }
            for(int v=w;v>=0;--v)
            {
                //跟上一轮比，得出v的最佳答案。
                dp[1][v] = max(dp[1][v],dp[0][v]);
                //滚动数组
                dp[0][v] = dp[1][v];
            }


        }

        printf("%d\n",dp[1][w]);
    }
    return 0;
}