HDU 2602 Bone Collector

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本文介绍了一道经典的背包问题——骨收集者问题。该问题是关于如何在有限的背包容量下，选择不同价值与体积的骨头以达到总价值最大化。文章通过示例详细解释了输入输出格式，并提供了一个使用动态规划解决此问题的C++实现代码。

Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 66486 Accepted Submission(s): 27737

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

Sample Output

思路：DP模板题

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>

using namespace std;

const double EXP = 1e-9;
typedef long long ll;

int main()
{
    int t,n,m;
    int dp[1005];
    int w[1005],v[1005];
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            scanf("%d%d",&n,&m);
            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=m;j>=w[i];j--)
                {
                    dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
                        //printf("dp[%d]=%d\n",j,dp[j]);
                }
            }
            printf("%d\n",dp[m]);
        }
    }
    return 0;
}