Bone Collector-HDU

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

思路：

 二维代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
long long max(long long num1,long long num2)//比较函数
{
return num1>num2?num1:num2;
}
long long dp[1005][1005];//定义背包数组
int main()
{
    int t;//数据组数
    int m_bone,m_volume;
    int bone[1005], volume[1005];
    cin >> t;
    while (t--)
    {
        int i, j, k;
        memset(dp, 0, sizeof(dp));//初始化
        memset(bone,0,sizeof(bone));
        memset(volume, 0, sizeof(volume));
        cin >> m_bone >> m_volume;//输入骨头总数和容量
        for (i = 1; i <= m_bone; i++)
            cin >> bone[i];//输入价值
        for (j = 1; j <= m_bone; j++)
            cin >> volume[j];//输入容量
        for (i = 1; i <= m_bone; i++)
        {
            for (j = 0; j <= m_volume; j++)
            {
                if (j < volume[i])//如果装不下
                {
                    dp[i][j] = dp[i - 1][j];//将上一个值赋给当前的价值
                    continue;
                }
                else
                {
                    dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - volume[i]] + bone[i]);//动态转移方程
                }
            }
        }
        cout << dp[m_bone][m_volume] << endl;//输出价值最大
    }
    return 0;
} 

优化成：一维代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
long long max(long long num1,long long num2)//比较函数
{
return num1>num2?num1:num2;
}
long long dp[1005];//定义背包数组
int main()
{
int N;
int i,j,k;
int bone[1005];//定义价值
int volume[1005];//定义容量
int t;
scanf("%d",&t);
while(t--)
{
memset(dp,0,sizeof(dp));//初始化
int count,weight;
scanf("%d%d",&count,&weight);//输入组数和总重量
for(j=1;j<=count;j++)
scanf("%d",&bone[j]);
for(j=1;j<=count;j++)
scanf("%d",&volume[j]);
for(i=1;i<=count;i++)
{
    for(k=weight;k>=0;k--)
{
        if(k-volume[i]<0)//装不下
        {
            break;
        }
        else
            dp[k]=max(dp[k-volume[i]]+bone[i],dp[k]);
        cout << dp[k] << endl;
    }
}
printf("%lld
",dp[weight]);
}
return 0;
}