HDU-2602-Bone Collector【01背包模板】

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Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 51862    Accepted Submission(s): 21839

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 2 ³¹).

 

Sample Input

  

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

 

Sample Output

  

   14
  

一维数组：

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int N,V;
int value[1010];
int volume[1010];
int dp[1010];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d %d",&N,&V);
		for(int i=0;i<N;i++)
			scanf("%d",&value[i]);
		for(int i=0;i<N;i++)
			scanf("%d",&volume[i]);
		memset(dp,0,sizeof(dp));
		for(int i=0;i<N;i++)
		{
			for(int j=V;j>=volume[i];j--)
			{
				dp[j]=max(dp[j],dp[j-volume[i]]+value[i]);
			}
		}
		printf("%d\n",dp[V]);
	}
	return 0;
}

二维数组：

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int N,V;
int val[1010];
int vol[1010];
int dp[1010][1010];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d %d",&N,&V);
		for(int i=1;i<=N;i++)
			scanf("%d",&val[i]);
		for(int i=1;i<=N;i++)
			scanf("%d",&vol[i]);
		memset(dp,0,sizeof(dp));
		for(int i=1;i<=N;i++)
		{
			for(int j=0;j<=V;j++)
			{
				dp[i][j]=dp[i-1][j];
				if(j>=vol[i])
				{
					dp[i][j]=max(dp[i-1][j],dp[i-1][j-vol[i]]+val[i]);
				}
			}
		/*	for(int j=vol[i];j<=V;j++)  错误的写法，当 j<vol[i] 的时候没有继承上一步的值 
			{
				dp[i][j]=max(dp[i-1][j],dp[i-1][j-vol[i]]+val[i]);
			}*/
		}
		printf("%d\n",dp[N][V]);
	}
	return 0;
}