背包九讲

1.01背包

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 85330    Accepted Submission(s): 35307
 

Problem Discription

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

 

 

#include
#include
#include
#include
#include
using namespace std;
int dp[1010][1010];
int volume[1010];
int value[1010];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
    {
        int n,v;
        scanf("%d%d",&n,&v);
        for(int i=1;i<=n;i++)
            scanf("%d",&value[i]);
        for(int i=1;i<=n;i++)
            scanf("%d",&volume[i]);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
            for(int j=0;j<=v;j++)
        {
            if(j

空间优化

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