dp--01背包--Charm Bracelet

Charm Bracelet

Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight W_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).

Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: W_i and D_i

Output

* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

Sample Input

4 6
1 4
2 6
3 12
2 7

Sample Output

23

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 dp[4000][12000];
 6 int w[1000],c[1000];
 7 int main()
 8 {
 9     int n,m;
10     scanf ("%d%d",&n,&m);
11     for (int i =1 ;i<= n;i++)
12     {
13         scanf ("%d%d",&w[i],&c[i]);
14     }
15     for (int i = 1;i <= n;i++)
16     {
17         for (int j = 0;j <= m;j++)
18         {
19             dp[i][j] = dp[i-1][j];
20             if (j-w[i]>=0)
21             {
22                 dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+c[i]);
23                 
24             }
25         
26         }    
27     //    cout<
28     }
29     cout<<dp[n][m];
30     return 0;
31 }

 

由这一道题的数据范围，显然二维数组不可实现，我们就把二维压缩为一维
所以这时，如果我们仍正序递推，每次下一个数据都有可能用到之前我们已经更新过的数据，所以我们应该从后开始

 1 #include
 2 using namespace std;
 3 int f[20000];
 4 int main()
 5 {
 6     int  n,v;
 7     int w[20000], c[20000];
 8     cin >> n >> v;
 9     for(int i = 1; i <= n; ++i)
10         cin >> w[i] >>c[i];
11     for(int i = 1; i <= n; ++i){
12         for (int j=v ;j>=w[i] ; j--){
13                 f[j] = max(f[j], f[j - w[i]] + c[i]);
14         }
15     }
16         cout << f[v]<<endl;
17     return 0;
18 }