dp--01背包--Bone Collector

Bone Collector

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

 

01背包：

首先我们要明白i，j，dp[i][j]都分别代表着什么

i:表示从1~i个物品里取

j:表示此时的最大容积

dp[i][j]:表示在1~i个物品里选取，每个物品只选择一次（限于01背包），且不超过j的容积，问所能得到的最大价值

得出一个转移方程f[i][j] = max(f[i][j], f[i - 1][j - c[i]] + w[i])

即在选取这个物品和不选这个物品中选择一个最大的

最后我们要输出的就是dp[n][m]表示在n个物品中，容积不超出m的最大价值

 

 1 #include
 2 #include 
 3 #include
 4 using namespace std;
 5 int f[1101][1101];
 6 int main()
 7 {
 8     int t;
 9     cin>>t;
10     for (int i=1 ;i<=t ;i++){
11         memset(f,0,sizeof f);
12         int  n,v;
13         int w[1001], c[1001];
14         cin >> n >> v;
15         for(int i = 1; i <= n; ++i)
16             cin >> w[i] ;
17         for (int i=1 ;i<=n ;i++)
18                cin>>c[i];
19         for(int i = 1; i <= n; ++i)
20             for(int j = 0; j <= v; ++j)
21             {
22                 f[i][j] = f[i - 1][j];
23                 if(j >= c[i])
24                     f[i][j] = max(f[i][j], f[i - 1][j - c[i]] + w[i]);
25             }
26         cout << f[n][v]<<endl;
27     }
28     return 0;
29 }