hdu 2602 Bone Collector

最新推荐文章于 2022-10-28 13:05:08 发布

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最新推荐文章于 2022-10-28 13:05:08 发布

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分类专栏： hdu water dp 文章标签： integer each output less ini input

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探讨了骨收集者在有限容量的包中收集不同价值和体积的骨头时，如何最大化总价值的问题，通过实例输入输出说明解决方案。

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题目描述：

Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14889 Accepted Submission(s): 5911

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

Sample Output

Author

Teddy

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

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lcy

思路：

最原始的背包问题：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<queue>

#define inf 0xfffffff
#define CLR(a,b) memset((a),(b),sizeof(a))

using namespace std;
int const nMax = 1100;

int dp[nMax];
int v[nMax],a[nMax];
int n,V;

int DP()
{
    CLR(dp,0);
    for(int k=0;k<n;k++){
        for(int i=V;i>=0;i--)if(i>=a[k]){
            dp[i]=max(dp[i],max(dp[i-1],dp[i-a[k]]+v[k]));
        }
    }
    return dp[V];
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&V);
        for(int i=0;i<n;i++)scanf("%d",&v[i]);
        for(int i=0;i<n;i++)scanf("%d",&a[i]);
        printf("%d\n",DP());
    }
    return 0;
}