Bone Collector--hdu2602（01背包）

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

 

Author

Teddy

 

Source

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

 

 

题意：给你n个骨头的价值和体积，求体积为V时最多有多少价值；

01背包问题:

递推公式：dp[i][j]=dp[i+1][j];(j<w[i]);

　　　　　dp[i][j]=max(dp[i+1][j],dp[i+1][j-w[i]]+v[i]);

代码如下：

 

关于i的逆向循环；

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
#define N 1100
   int dp[N][N];
int main()
{
    int T,i,j,V,n,v[N],w[N];

    scanf("%d",&T);
    while(T--)
    {
        memset(dp,,sizeof(dp));
        scanf("%d%d",&n,&V);
        for(i=;i<n;i++)
                scanf("%d",&v[i]);
        for(i=;i<n;i++)
                scanf("%d",&w[i]);
        for(i=n-;i>=;i--)
        {
            for(j=;j<=V;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]);
            }
        }
        printf("%d\n",dp[][V]);
    }
    return ;
}

下面是关于i的正向循环；

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
#define N 1100
   int dp[N][N];
int main()
{
    int T,i,j,V,n,v[N],w[N];

    scanf("%d",&T);
    while(T--)
    {
        memset(dp,,sizeof(dp));
        scanf("%d%d",&n,&V);
        for(i=;i<n;i++)
                scanf("%d",&v[i]);
        for(i=;i<n;i++)
                scanf("%d",&w[i]);
        for(i=;i<n;i++)
        {
            for(j=;j<=V;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]);
            }
        }
        printf("%d\n",dp[n][V]);
    }
    return ;
}

用一维数组；

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
#define N 1100
int dp[N];
int main()
{
    int T,i,j,V,n,v[N],w[N];

    scanf("%d",&T);
    while(T--)
    {
        memset(dp,,sizeof(dp));
        scanf("%d%d",&n,&V);
        for(i=;i<n;i++)
                scanf("%d",&v[i]);
        for(i=;i<n;i++)
                scanf("%d",&w[i]);
        for(i=;i<n;i++)
        {
            for(j=V;j>=w[i];j--)
            {
                dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
            }
        }
        printf("%d\n",dp[V]);
    }
    return ;
}