Charm bracelet-poj3624 (01 backpack)

http://poj.org/problem?id=3624

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 29444		Accepted: 13198

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


This is a simple 01 backpack.   
Start your backpack Journey
The normal backpack is
I     1....N
J     m ..... 0
Dp[i][j]=max (Dp[i-1][j],dp[i-1][j-w[i]]+v[i]);
But this range is bigger.
If you use two-dimensional hyper-memory
can be transformed into one-dimensional
Dp[j]=max (Dp[j],dp[j-w[i]]+v[i]);

#include <stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<algorithm>#include<iostream>using namespacestd;#defineN 3500intdp[13000];intMain () {intN,m,w[n],v[n];  while(SCANF ("%d%d", &n,&m)! =EOF) {         for(intI=1; i<=n;i++) scanf ("%d%d",&w[i],&V[i]); Memset (DP,0,sizeof(DP));  for(intI=1; i<=n;i++)        {             for(intj=m;j>=0; j--)            {                if(j>=W[i]) dp[j]=max (dp[j],dp[j-w[i]]+V[i]); }} printf ("%d\n", Dp[m]); }    return 0;}

Charm bracelet-poj3624 (01 backpack)