本文主要是介绍POJ 3624 Charm Bracelet 0/1背包基础，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目链接：POJ 3624 Charm Bracelet
B. Charm Bracelet
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: Wiand 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经典0/1背包问题
参看《背包九讲》
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;int n, m, dp[12890], w[3410], d[3410];
int main()
{scanf("%d%d", &n, &m);for(int i = 1; i <= n; i++)scanf("%d%d", &w[i], &d[i]);for(int i = 1; i <= n; i++)for(int j = m; j >= w[i]; j--)dp[j] = max(dp[j], dp[j-w[i]]+d[i]);printf("%d\n", dp[m]);return 0;
}这篇关于POJ 3624 Charm Bracelet 0/1背包基础的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
