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 2 ³¹).

 

 

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背包問題，這種背包特點是：每種物品僅有一件，可以選擇放或不放。
用子問題定義狀態：即dp[i][v]表示前i件物品恰放入一個容量為v的背包可以獲得的最大價值。
則其狀態轉移方程便是：
dp[i][v]=max{dp[i-1][v],dp[i-1][v-cost[i]]+value[i]}

注意體積為零的情況，如：
1
5 0
2 4 1 5 1
0 0 1 0 0
結果為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個物品將放入大小為j的背包中
                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-vo[i]]+va[i]);//第i個物品放入后，那么前i-1個物品可能會放入也可能因為剩余空間不夠無法放入
                else //第i個物品無法放入
                    dp[i][j]=dp[i-1][j];
            }
        }
        cout<<dp[n][v]<<endl;
    }
    return 0;
}

 

 

該題的第二種解法就是對背包的優化解法，當然只能對空間就行優化，時間是不能優化的。

先考慮上面講的基本思路如何實現，肯定是有一個主循環i=1..N，每次算出來二維數組dp[i][0..V]的所有值。
那么，如果只用一個數組dp[0..V]，能不能保證第i次循環結束后dp[v]中表示的就是我們定義的狀態dp[i][v]呢？

dp[i][v]是由dp[i-1][v]和dp[i-1][v-c[i]]兩個子問題遞推而來，能否保證在推dp[i][v]時（也即在第i次主循環中推dp[v]時）能夠得到dp[i-1][v]和dp[i-1][v-c[i]]的值呢？事實上，這要求在每次主循環中我們以v=V..0的順序推dp[v]，這樣才能保證推dp[v]時dp[v-c[i]]保存的是狀態dp[i-1][v-c[i]]的值。偽代碼如下：

for i=1..N

    for v=V..0

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

注意：這種解法只能由V--0，不能反過來，如果反過來就會造成物品重復放置！

 

#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;
}