HDU 2602 Bone Collector（01背包）

 

 

          Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14336    Accepted Submission(s): 5688

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

 

 

Recommend

lcy

 

 

 

01背包问题，这种背包特点是：每种物品仅有一件，可以选择放或不放。
用子问题定义状态：即dp[i][v]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值。
则其状态转移方程便是：
dp[i][v]=max{dp[i-1][v],dp[i-1][v-cost[i]]+value[i]}

注意体积为零的情况，如：
1
5 0
2 4 1 5 1
0 0 1 0 0
结果为12

 

#include<iostream>
using namespace std;
int dp[1000][1000];

int max(int x,int y)
{
    return x>y?x:y;
}

int main()
{
    int t,n,v,i,j;
    int va[1000],vo[1000];
    cin>>t;
    while(t--)
    {
        cin>>n>>v;
        for(i=1;i<=n;i++)
            cin>>va[i];
        for(i=1;i<=n;i++)
            cin>>vo[i];
        memset(dp,0,sizeof(dp));//初始化操作
         for(i=1;i<=n;i++)
        {
            for(j=0;j<=v;j++)
            {
                if(vo[i]<=j)//表示第i个物品将放入大小为j的背包中
                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-vo[i]]+va[i]);//第i个物品放入后，那么前i-1个物品可能会放入也可能因为剩余空间不够无法放入
                else //第i个物品无法放入
                    dp[i][j]=dp[i-1][j];
            }
        }
        cout<<dp[n][v]<<endl;
    }
    return 0;
}

 

 

该题的第二种解法就是对背包的优化解法，当然只能对空间就行优化，时间是不能优化的。

先考虑上面讲的基本思路如何实现，肯定是有一个主循环i=1..N，每次算出来二维数组dp[i][0..V]的所有值。
那么，如果只用一个数组dp[0..V]，能不能保证第i次循环结束后dp[v]中表示的就是我们定义的状态dp[i][v]呢？

dp[i][v]是由dp[i-1][v]和dp[i-1][v-c[i]]两个子问题递推而来，能否保证在推dp[i][v]时（也即在第i次主循环中推dp[v]时）能够得到dp[i-1][v]和dp[i-1][v-c[i]]的值呢？事实上，这要求在每次主循环中我们以v=V..0的顺序推dp[v]，这样才能保证推dp[v]时dp[v-c[i]]保存的是状态dp[i-1][v-c[i]]的值。伪代码如下：

for i=1..N

    for v=V..0

        dp[v]=max{dp[v],dp[v-c[i]]+w[i]};

注意：这种解法只能由V--0，不能反过来，如果反过来就会造成物品重复放置！

 

#include<iostream>
using namespace std;
#define Size 1111
int va[Size],vo[Size];
int dp[Size];
int Max(int x,int y)
{
    return x>y?x:y;
}
int main()
{
    int t,n,v;
    int i,j;
    cin>>t;
    while(t--)
    {
        cin>>n>>v;
        for(i=1;i<=n;i++)
            cin>>va[i];
        for(i=1;i<=n;i++)
            cin>>vo[i];
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
        {
            for(j=v;j>=vo[i];j--)
            {
                dp[j]=Max(dp[j],dp[j-vo[i]]+va[i]); 
            }
        }
        cout<<dp[v]<<endl;
    }
    return 0;
}