poj-3624 Charm Bracelet 01背包
- 2019 年 10 月 10 日
- 筆記
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 54588 | Accepted: 22773 |
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 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: Wi and 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
Source
poj-3624
author：
Caution_X
date of submission：
20191006
tags：
01背包+空间优化
description modelling：
01背包
major steps to solve it：
1.dp[i]:=容量为i时的最大价值
2.dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
warnings：
不使用空间优化会MLE
#include<cstdio> #include<algorithm> using namespace std; int w[3500],v[3500]; int dp[12881]; int main() { int n,k; scanf("%d%d",&n,&k); for(int i=0;i<n;i++) { scanf("%d%d",&w[i],&v[i]); } for(int i=0;i<n;i++) { for(int j=k;j>=w[i];j--) { dp[j]=max(dp[j],dp[j-w[i]]+v[i]); } } printf("%dn",dp[k]); return 0; }
