HDU-Bone Collector（01背包问题）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 49837    Accepted Submission(s): 20878

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

 

Author

Teddy

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

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典型的01背包问题

#include
#include
#include
#include
#include
using namespace std;
int dp[1005][1005],a[1005],b[1005];
int  main()
{
	int t,n,m,i,j;
	scanf("%d",&t);
	while(t--)
	{
		memset(dp,0,sizeof(dp));
		scanf("%d%d",&n,&m);
		for(i=1;i<=n;i++)
			scanf("%d",&a[i]);
		for(i=1;i<=n;i++)
			scanf("%d",&b[i]);
		for(i=1;i<=n;i++)
			  for(j=0;j<=m;j++)
			  {
				  if(b[i]<=j)
					  dp[i][j]=max(dp[i-1][j],dp[i-1][j-b[i]]+a[i]);
				  else
					  dp[i][j]=dp[i-1][j];
			  }
		printf("%d\n",dp[n][m]);
	}
}

优化代码：

#include
#include
#include
#include
#include
using namespace std;
int dp[1005],a[1005],b[1005];
int  main()
{
int t,n,m,i,j;
scanf("%d",&t);
while(t--)
{
memset(dp,0,sizeof(dp));
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
for(i=1;i<=n;i++)
scanf("%d",&b[i]);
for(i=1;i<=n;i++)
for(j=m;j>=b[i];j--)
dp[j]=max(dp[j-b[i]]+a[i],dp[j]);
printf("%d\n",dp[m]);
}
}