Bone Collector

布满荆棘的人生 2022-08-18 02:26 35阅读 0赞

## Bone Collector ##

####  ####

##### Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^\_^ #####

## 题目描述 ##

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 ?

## 输入 ##

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.

## 输出 ##

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

## 示例输入 ##

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

## 示例输出 ##

## 提示 ##

[hdoj2602][]

## 来源 ##

## 示例程序 ##

#include <stdio.h>
        #define MAXN 10002
        int max_num(int a, int b){
            return a > b? a : b;
        }
        int dp[MAXN], w[MAXN], p[MAXN];
        int main(){
    
            int n, v, i, j,t;
            scanf("%d",&t);
            while(t--)
                {
                    scanf("%d %d", &n, &v);
                for(i=1; i<=n; i++)
                    {
                    scanf("%d",&w[i]);
                    }
                    for(i=1; i<=n; i++)
                    {
                    scanf("%d",&p[i]);
                    }
                  for(i=0; i<=v; i++){
                    dp[i] = 0;
                }
                for(i=1; i<=n; i++){
                    for(j=v; j>=0; j--){
                        if(j>=p[i]) dp[j] = max_num(dp[j], dp[j-p[i]]+w[i]);
                    }
                }
                printf("%d\n", dp[v]);
                }
    
        }

[hdoj2602]: http://acm.hdu.edu.cn/showproblem.php?pid=2602