hdu 2602 Bone Collector

最新推荐文章于 2022-10-28 13:05:08 发布

原创最新推荐文章于 2022-10-28 13:05:08 发布 · 244 阅读

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本文针对骨收集者问题，采用动态规划方法解决背包问题。在限定背包容量的情况下，选取不同价值与体积的骨头以获得最大总价值。文章通过具体实例进行讲解，并提供了完整的代码实现。

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Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 61757 Accepted Submission(s): 25729

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 2 ³¹).

Sample Input

Sample Output

就是用动态规划，推荐一个讲的不错的博客，看懂他在表达什么意思了，然后自己推一遍，然后这个题迎刃而解。下面是我的代码！

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[1005];
int b[1005];
int dp[1005][1005];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(dp,0,sizeof(dp));
        int i,n,v,j;
        scanf("%d%d",&n,&v);
        for(i=1; i<=n; i++)
            scanf("%d",&a[i]);
        for(i=1; i<=n; i++)
            scanf("%d",&b[i]);
        for(i=0; i<=v; i++)
        {
            if(i<b[n])
                dp[n][i]=0;
            else
                dp[n][i]=a[n];
        }
        for(i=n-1; i>=1; i--)
        {
            for(j=0; j<=v; j++)
            {
                if(j<b[i])
                    dp[i][j]=dp[i+1][j];
                else
                    dp[i][j]=max(dp[i+1][j],dp[i+1][j-b[i]]+a[i]);
            }
        }
        printf("%d\n",dp[1][v]);
        /* for(i=0;i<=n;i++)
         {
             for(j=0;j<=v;j++)
                 printf("%d ",dp[i][j]);
             printf("\n");
         }*/
    }
    return 0;
}