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POJ 3624 Charm Bracelet C++版

POJ题解 Slyar 4233浏览 1评论

文章作者:姜南(Slyar) 文章来源:Slyar Home (www.slyar.com) 转载请注明,谢谢合作。

DP 01背包

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

Slyar:标准的01背包,动态转移方程如下。其中dp[i,j]表示的是前i个物品装入容量为j的背包里所能产生的最大价值,w[i]是第i个物品的重量,d[i]是第i个物品的价值。如果某物品超过背包容量,则该物品一定不放入背包,问题变为剩余i-1个物品装入容量为j的背包所能产生的最大价值;否则该物品装入背包,问题变为剩余i-1个物品装入容量为j-w[i]的背包所能产生的最大价值加上物品i的价值d[i]…恩,多清晰…

还有一点是,这题我之前使用二维数组做的,不过提交之后MLE了…所以只能想办法使用一维数组,手工模拟一下发现dp[i,j]只跟之前的某几个值有关,再之前计算的结果就都没用了…所以我们就可以用一个数组来重复使用了,不过注意要倒序,保证无后效性…

if (w[i]>j) dp[i,j] = dp[i-1,j]

else dp[i,j] = max(dp[i-1,j-w[i]]+d[i],dp[i-1,j])

#include 

using namespace std;

const int N = 3403;
const int M = 12881;

int dp[M] = {0};

int max(int x, int y)
{
	return x > y ? x : y;
}

int main()
{
	int n, m;
	int w[N], d[M];

	cin >> n >> m;

	for (int i = 1; i <= n; i++)
	{
		cin >> w[i] >> d[i];
	}

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

	cout << dp[m] << endl;

	return 0;
}
所能产生的最大价值

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网友最新评论 (1)

  1. 请教下: for (int j = m; j >= w[i]; j--) { dp[j] = max(dp[j-w[i]] + d[i], dp[j]); } 这个循环中为什么j不从小到大呢
    匿名9年前 (2012-05-15)回复