POJ3624 Charm Bracelet

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15656		Accepted: 7120

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 theN (1 ≤N ≤ 3,402) available charms. Each charmi in the supplied list has a weightW_i (1 ≤W_i ≤ 400), a 'desirability' factorD_i (1 ≤D_i ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more thanM (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 andD_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

Source

USACO 2007 December Silver


背包加空间压缩
具体代码：

#include<iostream>
#define SIZE 3500
#define SIZE2 12882
using namespace std;
struct node
{
	int w;
	int d;
};
int w[SIZE],d[SIZE],g[SIZE2];

int min(int x1, int x2)
{
	if(x1 < x2)
		return x1;
	else
		return x2;
}

int max(int x1, int x2)
{
	if(x1 > x2)
		return x1;
	else
		return x2;
}
int main()
{
	int m,n,i,j;
	long res;

	//input
	cin >> n >> m;
	for(i = 1; i <= n; i++)
		cin >> w[i] >> d[i];
	//init
	for(j = 1; j <= m; j++)
		g[j] = 0;

	for(i = 1; i <= n; i++)
	{
		for(j = m; j >= w[i]; j--)
		{
			g[j] = max(g[j],g[j - w[i]] + d[i]);
		}
	}

	//output
	res = g[m];
	cout << res << endl;

	return 0;
}