牛客假日团队赛6 H：Charm Bracelet （01背包）

链接：https://ac.nowcoder.com/acm/contest/993/H
来源：牛客网
 

题目描述

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.

输入描述:

* 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

输出描述:

* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

示例1

输入

复制

4 6
1 4
2 6
3 12
2 7

输出

复制

23

题意分析：

Bessie有m点魅力，现在有n个项链，每个项链会消耗魅力来获得价值，求能够获得的最大价值

解题思路：

01背包

#include <stdio.h>
#include <algorithm>
using namespace std;
#define N 15000
int dp[N], w[N], d[N];
int main()
{
	int n, m, i, j;
	scanf("%d%d", &n, &m);
	for (i = 1; i <= n; i++)
		scanf("%d%d", &w[i], &d[i]);
	for (i = 1; i <= n; i++)
		for (j = m; j >= w[i]; j--)
			dp[j] = max(dp[j], dp[j - w[i]] + d[i]);
	printf("%d
", dp[m]);
	return 0;
}