HDU 3624 Charm Bracelet （01背包）

Charm Bracelett

Time Limit: 1000MS
Memory Limit: 65536KTotal Submissions: 13977
Accepted: 6381

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

4 61 42 63 122 7

Sample Output

23

Source

USACO 2007 December Silver

 

 

 

简单的01背包，

开始练习背包

 

 

/*简单的01背包*/#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>using namespace std;int dp[13000];//dp【i】表示当重量为i时  它最大价值 int n,w;int we[10000];//每袋的重量 int v[10000];//每袋的价值 int main(){while(cin>>n>>w){for(int i=0;i<n;i++)cin>>we[i]>>v[i];memset(dp,0,sizeof(dp));for(int i=0;i<n;i++)for(int j=w;j>=we[i];j--)dp[j]=max(dp[j],dp[j-we[i]]+v[i]);cout<<dp[w]<<endl;}return 0;}