POJ 3624 Charm Bracelet (backpack)

Charm Bracelet Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 24914 Accepted: 11226 Description Bessie have 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 Char Ms. Each charm I in the supplied list has a weight wi (1≤ wi ≤400), a ' desirability ' factor C5>di (1≤ Di ≤100), and can be used at the most once. Bessie can only support a charm bracelet whose weight are 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 MA Ximum 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 II space-separated integers: Wi and Di Output * Line 1: A single integer which is the greatest sum of charm desirabilities so can be achieved given the weight Constrai Nts Sample Input 4 61 42 63 122 7 Sample Output 23 Source Usaco December Silver

Naked Backpack.

Code:

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm>using namespace Std;const int maxn=30000+1000;int Dp[maxn];int main () {    int n,m,w,v;    while (~SCANF ("%d%d", &n,&m))    {      memset (dp,0,sizeof (DP));      for (int i=0;i<n;i++)      {          scanf ("%d%d", &w,&v);          for (int j=m;j>=w;j--)          {              Dp[j]=max (dp[j-w]+v,dp[j]);          }      }      int ans=0;      for (int i=0;i<=m;i++)      Ans=max (Ans,dp[i]);      printf ("%d\n", ans);    }    return 0;}

POJ 3624 Charm Bracelet (backpack)