杭电-2602 Bone Collector（01背包模板）-优快云博客

本文链接：https://blog.youkuaiyun.com/lu_1110/article/details/52209527

本文介绍了一个经典的01背包问题实例——骨收集者问题。该问题要求根据给定的骨头价值和体积，以及骨收集者的背包容量，计算出骨收集者能够获得的最大总价值。文章提供了完整的代码实现，并详细解释了输入输出格式。

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Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 51853 Accepted Submission(s): 21833

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

Author

Teddy

Source

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

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01背包：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int v[1100],w[1100],dp[1100];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int N,V;
        scanf("%d%d",&N,&V);
        for(int i = 1; i <= N; i++)
            scanf("%d",&v[i]);
        for(int i = 1; i <= N; i++)
            scanf("%d",&w[i]);
        memset(dp,0,sizeof(dp));
        for(int i = 1; i <= N; i++)
        {
            for(int j = V; j >= w[i]; j--)
            {
                dp[j] = max(dp[j],dp[j-w[i]] + v[i]);
            }
        }
        printf("%d\n",dp[V]);
    }
    return 0;
}