hdu 2602 基础01背包

最新推荐文章于 2022-09-09 22:07:37 发布

charote233

最新推荐文章于 2022-09-09 22:07:37 发布

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分类专栏：背包

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本文介绍了一道经典的01背包问题，通过一个骨收集者的故事背景来阐述如何利用动态规划算法求解最大价值的问题。文章提供了完整的代码实现，并对输入输出格式及样例进行了详细说明。

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Problem Description
Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

Input
The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

Output
One integer per line representing the maximum of the total value (this number will be less than 231).

Sample Input
1
5 10
1 2 3 4 5
5 4 3 2 1

Sample Output
14

题意：基础01背包，做的第一个模板题，直接放代码了

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[1050],b[1050],dp[1010][1010];

int main()
{
    int t,n,k;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        scanf("%d%d",&n,&k);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&b[i]);
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=k;j>=0;j--)
            {
                if(j>=b[i])
                  dp[i][j]=max(dp[i-1][j],dp[i-1][j-b[i]]+a[i]);
                else
                  dp[i][j]=dp[i-1][j];
            }
        }
        printf("%d\n",dp[n][k]);
    }
    return 0;
}