Charm Bracelet

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#DP #背包 #动态规划_背包

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本文介绍了一种典型的背包问题——魅力手链问题。该问题旨在寻找在重量限制下能达到的最大魅力值组合。通过动态规划算法解决，输入包括魅力的数量及其各自的重量与魅力值，输出为最大总魅力值。

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Charm Bracelet

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 91 Accepted Submission(s) : 45

Problem 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

Sample Output

Source

PKU

Statistic | Submit | Back

#include<stdio.h>
#include<string.h>
#define NN 3402+5
#define MM 12880+5
int cost[NN], value[NN], dp[MM];
int max(int a, int b)
{
	return a > b ? a : b;
}
int main()
{
	int N, M, i, j;
	memset(dp, 0, sizeof(dp));
	scanf("%d%d", &N, &M);
	for (i = 1; i <= N; i++)
		scanf("%d%d", &cost[i], &value[i]);
	for (i = 1; i <= N; i++)
		for (j = M; j >=0; j--){
			if (j >= cost[i])
				dp[j] = max(dp[j], dp[j - cost[i]] + value[i]);
		}
	printf("%d\n", dp[M]);
	return 0;
}