POJ 3624 Charm Bracelet (01背包 + 空间优化)

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 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

#include 
#include 
#include 
using namespace std;
#define N 3500
#define M 13000
int w[N], d[N];
int dp[M];
int n, m;
int main()
{
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) scanf("%d%d", &w[i], &d[i]);
	memset(dp, 0, sizeof(dp));
	for (int i = 1; i <= n; i++)
		for (int j = m; j >= w[i]; j--)
		{
			dp[j] = max(dp[j], d[i] + dp[j - w[i]]);
		}
	printf("%d\n", dp[m]);
	return 0;
}