dp重拾-01背包--HDU 2602

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

 

挺久没刷DP题了，先来一个01背包预热一下

#include<iostream>
#include<cstring>
#include<stdlib.h>
#include<stdio.h>
using namespace std;
int dp[][];
int w[],v[],W,n;
int max(int a,int b)
{
    if(a>=b) return a;
    else return b;
}
void solve()
{
    for(int i=n-;i>=;i--)
    {
        for(int j=;j<=W;j++)
        {
            if(j<w[i]) dp[i][j]=dp[i+][j];
            else dp[i][j]=max(dp[i+][j],dp[i+][j-w[i]]+v[i]);
        }
    }
    cout<<dp[][W]<<endl;
}

int main()
{
    int i,j,k;
    scanf("%d",&k);
    for(i=;i<k;i++)
    {
        scanf("%d %d",&n,&W);
        for(j=;j<n;j++)
        {
            scanf("%d",&v[j]);
        }
        for(j=;j<n;j++)
        {
            scanf("%d",&w[j]);
        }
        solve();
        W=n=;
        memset(dp,,sizeof(dp));
        memset(v,,sizeof(v));
        memset(w,,sizeof(w));
    }
    return ;
}