POJ 3624 Charm Bracelet (01背包 O（v)空间复杂度）

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions:42278		Accepted: 18245

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

• Source Code

  #include<iostream>
  #include<algorithm>
  #include<string.h>
  #include<cstdio>
  int dp[13000];
  using namespace std;
  int main()
  {
  	int N, M;
  	while (scanf_s("%d%d", &N, &M) != EOF)
  	{
  		memset(dp, 0, sizeof(dp));
  		int n[3500][2];
  		for (int i = 1; i <= N; i++)
  			scanf_s("%d%d", &n[i][0], &n[i][1]);
  		for (int i = 1; i <= N; i++)
  		{
  			for (int j = M; j >= 1; j--)
  			{
  				if (j >= n[i][0])
  					dp[j] = max(dp[j], dp[j - n[i][0]] + n[i][1]);
  			}
  		}
  		printf("%d\n", dp[M]);
  	}
  	return 0;
  }