Bone Collector hdu 2602 c++ 01背包基础题

柔光的暖阳◎ 2022-08-10 15:51 131阅读 0赞

# Bone Collector #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 29278    Accepted Submission(s): 11984**

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 ?  
![C154-1003-1.jpg][]

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

1
    5 10
    1 2 3 4 5
    5 4 3 2 1

Sample Output

#include<cstdio>
    #include<cstring>
    #define max(a,b)  (a>b?a:b)
    #define MAX 2000
    using namespace std;
    int va[MAX],vo[MAX],f[MAX];
    
    int main()
    {
        int i,j,t,m,n;
        scanf("%d",&n);
        while(n--)
        
        {
            scanf("%d%d",&t,&m);
            memset(va,0,sizeof(va));
            memset(vo,0,sizeof(vo));
            memset(f,0,sizeof(f));
            for(i=1;i<=t;i++)
                scanf("%d",&va[i]);
            for(i=1;i<=t;i++)  
                scanf("%d",&vo[i]);  
            for(i=1;i<=t;i++)  
                for(j=m;j>=vo[i];j--)
                    f[j]=max(f[j],f[j-vo[i]]+va[i]);
            printf("%d\n",f[m]);        
                        
        } 
        return 0;
    }

**f\[i\]\[v\]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值。则其状态转移方程便是：**

**f\[i\]\[v\]=max\{f\[i-1\]\[v\],f\[i-1\]\[v-c\[i\]\]+w\[i\]\}**

[C154-1003-1.jpg]: /images/20220810/ea9ca15aa6ec43208985b6ea09f5b34c.png