Poj3624 Charm Bracelet

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字数统计: 355 | 阅读时长 ≈ 2

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

1
2
3
4
5
4 6
1 4
2 6
3 12
2 7

Sample Output

1
23

Analysis

N 个物品每个物品有价值d[i]，重量w[i]， 给定背包最大承重M，求背包能够装载的最大价值。每个物品只有放入背包和不放入背包两种选择。这是典型的0-1背包问题。

直接套用模板即可。

Code

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#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
int N, M;
int dp[12800 + 1];
int W[3402 + 1];
int D[3402 + 1];
int main()
{
	while (cin >> N >> M)
	{
		memset(dp, 0, sizeof(dp));
		memset(W, 0, sizeof(W));
		memset(D, 0, sizeof(D));
		for (int i = 0; i < N; i++)
		{
			cin >> W[i] >> D[i];
			for (int j = M; j >= W[i]; j--)
			{
				dp[j] = max(dp[j], dp[j - W[i]] + D[i]);
			}
		}
		cout << dp[M] << endl;
	}
	return 0;
}