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.

有 �N 件物品和一个容量为 �M 的背包。第 �i 件物品的重量是 ��Wi，价值是 ��Di。求解将哪些物品装入背包可使这些物品的重量总和不超过背包容量，且价值总和最大。

输入格式

* 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

第一行：物品个数 �N 和背包大小 �M。

第二行至第 �+1N+1 行：第 �i 个物品的重量 ��Wi 和价值 ��Di。

输出格式

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

输出一行最大价值。

输入输出样例

输入 #1

4 6

1 4

2 6

3 12

2 7

输出 #1

23

代码如下：

#include<cstdio>
int n,m; 
int w[3500]={},v[3500]={},dp[20001]={};
int main(){scanf("%d%d",&n,&m);  for(int i=1;i<=n;i++) scanf("%d%d",&w[i],&v[i]);  for(int i=1;i<=n;i++){for(int j=m;j>=w[i];j--){if(dp[j-w[i]]+v[i]>dp[j]) dp[j]=dp[j-w[i]]+v[i];}}printf("%d",dp[m]);
}