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本文介绍了一个经典的背包问题求解案例，通过动态规划算法实现。案例中骨骸收集者需选择不同价值与体积的骨骸装入固定容量的背包内，以使总价值最大化。文章详细展示了算法流程及核心代码。

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#include <iostream>
#include<stdio.h>
#include <algorithm>
#include<cstring>
using namespace std;
int c[1005], v[1005], f[1005][1005];
int main()
{
    int i, j, n, m, t;
    cin>>t;
    while(t--)
    {
        memset(c, 0, sizeof(c));
        memset(v, 0, sizeof(v));
        memset(f, 0, sizeof(f));
        cin>>n>>m;
        for(i=1; i<=n; i++)
            cin>>v[i];
        for(i=1; i<=n; i++)
            cin>>c[i];

        for(i=1; i<=n; i++)
        {
            for(j=0; j<=m; j++)
            {
                if(c[i] > j)
                    f[i][j] = f[i-1][j];
                else
                    f[i][j] =  max(f[i-1][j], f[i-1][j-c[i]]+v[i]);
            }
        }
        cout<<f[n][m]<<endl;
    }
    return 0;
}

A - Bone Collector
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit

Status
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

5 10
1 2 3 4 5
5 4 3 2 1
Sample Output