01背包（就地滚动）

<span style="color:#3333ff;">/*
__________________________________________________________________________________________________
*     copyright:   Grant Yuan                                                                     *
*     algorithm:   01背包（就地滚动）                                                             *
*     time     :   2014.7.18                                                                      *
*     declare  :   题目中说N最大是3400多。可是一開始开了5000内存还是执行时错误，后来直接改了50000 *                                                       *                                                                                                 *
*_________________________________________________________________________________________________*</span>

<span style="color:#3333ff;">
I - 01背包（就地滚动）
Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
Submit
 
Status
Description
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 Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 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: Wi and Di

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
*/

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

int w[50001];
int p[50001];
int sum;
int n;
int dp[50001];

int main()
{
	cin>>n>>sum;
	for(int i=0;i<n;i++)
		cin>>w[i]>>p[i];
	memset(dp,0,sizeof(dp));
	for(int i=0;i<n;i++)
	   for(int j=sum;j>=w[i];j--)
	    {
		dp[j]=max(dp[j],dp[j-w[i]]+p[i]);
	     }
	cout<<dp[sum]<<endl;
	return 0;
}
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