题目：
Bone Collector
**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 53983 Accepted Submission(s): 22616
**
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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 ?\
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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.
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Output
One integer per line representing the maximum of the total value (this number will be less than 2 31).
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Sample Input
1
5 10
1 2 3 4 5
5 4 3 2 1
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Sample Output
14
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Author
Teddy
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Source
HDU 1st “Vegetable-Birds Cup” Programming Open Contest
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题意：
意思是一个人有收集骨头的习惯，现在他有一个背包，面前有不同的骨头种类和各自对应的价值，求这个背包能装的最大价值，这些东西不可分割，所以不能用贪心，也就是01背包问题
代码一（一维）：
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a));
using namespace std;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,v;//个数 和 容量
scanf("%d %d",&n,&v);
int a[1010],b[1010];
for(int i=0; i<n; i++)
scanf("%d",&a[i]);//每一个的价值
for(int i=0; i<n; i++)
scanf("%d",&b[i]);//每一个的体积
int dp[1010];//定义状态
mem(dp,0);
for(int i=0; i<n; i++)
{
for(int j=v; j>=b[i]; j--)//逆序
dp[j]=max(dp[j],dp[j-b[i]]+a[i]);
}
printf("%d\n",dp[v]);
}
return 0;
}
//dp[i][v]是由dp[i-1][v]和dp[i-1][v-c[i]]两个子问题递推而来，能否保证在推dp[i][v]时（也即在第i次主循环中推dp[v]时）能够得到dp[i-1][v]和dp[i-1][v-c[i]]的值呢？事实上，这要求在每次主循环中我们以v=V..0的顺序推dp[v]，这样才能保证推dp[v]时dp[v-c[i]]保存的是状态dp[i-1][v-c[i]]的值。伪代码如下：
// for i=1..N
// for v=V..0
// dp[v]=max{dp[v],dp[v-c[i]]+w[i]};
//注意：这种解法只能由V--0，不能反过来，如果反过来就会造成物品重复放置！
代码二（二维）：
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a));
using namespace std;
int dp[1000][1000];//定义状态
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,v;//个数 和 容量
scanf("%d %d",&n,&v);
int a[1010],b[1010];
for(int i=1; i<=n; i++)
scanf("%d",&a[i]);//每一个的价值
for(int i=1; i<=n; i++)
scanf("%d",&b[i]);//每一个的体积
mem(dp,0);
for(int i=1; i<=n; i++)//遍历个数
{
for(int j=0; j<=v; j++)//遍历背包容量
{
if(b[i]<=j)//可以放进去
dp[i][j]=max(dp[i-1][j],dp[i-1][j-b[i]]+a[i]);//第i个物品放入后,那么前i-1个物品可能会放入也可能因为剩余空间不够无法放入
else//不能放进去
dp[i][j]=dp[i-1][j];
}
}
printf("%d\n",dp[n][v]);
}
return 0;
}
ps:
01背包问题，这种背包特点是：每种物品仅有一件，可以选择放或不放。
用子问题定义状态：即dp[i][v]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值。
则其状态转移方程便是：
dp[i][v]=max{dp[i-1][v],dp[i-1][v-cost[i]]+value[i]}
另NYOJ289 苹果和这道题很相似
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