hdu 2602 （01背包）

Bone Collector

 

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

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
using namespace std;
#define ll long long 
const int mod=1e9+7;
const int inf=1e9+7;
const int maxn=1005;
int dp[maxn][maxn];
int value[maxn];
int capacity[maxn];
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int T;
    cin>>T;
    while(T--)
    {
        memset(dp,0,sizeof(dp));
        int n,v;
        cin>>n>>v;
        for(int i=1;i<=n;i++)
            cin>>value[i];
        for(int i=1;i<=n;i++)
            cin>>capacity[i];
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<=v;j++)//注意体积为0时的情况 
            {
                if(j<capacity[i])//放不进 
                    dp[i][j]=dp[i-1][j];
                else
                {
                    if(dp[i-1][j]>dp[i-1][j-capacity[i]]+value[i])//放得进但是亏了 
                        dp[i][j]=dp[i-1][j];
                    else
                        dp[i][j]=dp[i-1][j-capacity[i]]+value[i];//不亏 
                }
            }
           }
        cout<<dp[n][v]<<endl;  
    }
    return 0;
}

 

 

一维优化：

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
using namespace std;
#define ll long long 
const int mod=1e9+7;
const int inf=1e9+7;
const int maxn=1005;
int dp[maxn];
int value[maxn];
int capacity[maxn];
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int T;
    cin>>T;
    while(T--)
    {
        memset(dp,0,sizeof(dp));
        int n,v;
        cin>>n>>v;
        for(int i=1;i<=n;i++)
            cin>>value[i];
        for(int i=1;i<=n;i++)
            cin>>capacity[i];
        for(int i=1;i<=n;i++)
        {
            for(int j=v;j>=capacity[i];j--)
            {
                //if(j>=capacity[i])//由j的循环条件知，j一定大于等于capacity[i] 
                dp[j]=max(dp[j],dp[j-capacity[i]]+value[i]);
            }
        }
        cout<<dp[v]<<endl;  
    }
    return 0;
}