HDU-2602，Bone Collector（01背包）

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：

 14

解题思路： 

这是个典型的01背包问题，状态转移方程模板如下：
 

for(int i=1;i<=n;i++)
{
    for(int j=V;j>=v[i];j--)
    {
	dp[j]=max(dp[j],dp[j-v[i]]+d[i]);
    }
}

程序代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX=1001;
int dp[MAX],d[MAX],v[MAX];
int t,n,V;
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		memset(dp,0,sizeof(dp));
		scanf("%d %d",&n,&V);//n代表骨头的总数，V代表袋子的容积 
		for(int i=1;i<=n;i++)
			scanf("%d",&d[i]);//每个骨头的价值 
		for(int j=1;j<=n;j++)
			scanf("%d",&v[j]);//每个骨头的容积 
		for(int i=1;i<=n;i++)
		{
			for(int j=V;j>=v[i];j--)
			{
				dp[j]=max(dp[j],dp[j-v[i]]+d[i]); 
			}//容积减去相应的骨头的容积，同时加上这个骨头的价值 
		}
		printf("%d\n",dp[V]);
	}
	return 0;
}