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 2 31).
样例：
Sample Input
1 5 10 1 2 3 4 5 5 4 3 2 1
Sample Output
14题目大意：有一个人喜欢收集骨头（呃），现在有N根骨头，每根骨头对应着各自的体积和价值，这个人的背包容量为V，求能装的骨头的最大总价值。代码：
#include<iostream>
#include<stdio.h>
#define max(a,b) a>b?a:b
struct bone{
int wei,val;
};
int f[1002][1002];
bone temp[1005];
int value(int num,int v){
for(int i=1;i<=num;++i){
for(int j=0;j<=v;++j){
if(j<temp[i-1].wei)
f[i][j]=f[i-1][j];
else
f[i][j]=max(f[i-1][j-temp[i-1].wei]+temp[i-1].val,f[i-1][j]);
}
}
return f[num][v];
}
using namespace std;
int main(){
int n,num,v;
scanf("%d",&n);
for(int i=0;i<n;++i){
scanf("%d%d",&num,&v);
for(int j=0;j<num;++j) scanf("%d",&temp[j].val);
for(int j=0;j<num;++j) scanf("%d",&temp[j].wei);
printf("%d\n",value(num,v));
}
return 0;
}
这是一道0-1背包问题，要解决的问题就是要放哪些物品进背包，总价值是多少。
而对于每一个物品只有放与不放两种情况：
1. 第i件放进去：价值为：f[i-1][v-temp[i-1].wei]+temp[i-1].val（背包容量减去第i件物品的体积，再加上对前i-1件进行判断后得到价值的最大值（容量为v-temp[i-1].wei的背包））
2. 第i件不放进去：价值为：f[i-1][v]（容量不减少（为v），对前i-1件进行判断后得到价值的最大值）
状态转移方程为：f[i][v] = max(f[i-1][v], f[i-1][v-w[i]]+c[i])
随后对两种情况取较大值。第一行初始化为0.从第一个物品开始列表记录背包容量内剩余不同容积存放前i个物品情况的最大值。
这个解法比较容易理解，下面的链接有对背包问题更详尽的解释，受启发于此：
https://www.cnblogs.com/A-S-KirigiriKyoko/p/6036368.html
https://www.cnblogs.com/xym4869/p/8513801.html
（其中的代码将第一列也初始化为0，这样子的话当物品中同时存在体积=背包容量的物品和体积=0的物品时就会出错，因此稍作改动）
原文地址：https://www.cnblogs.com/-y-i-/p/10206060.html