POJ 3624 Charm Bracelet

最新推荐文章于 2019-08-06 15:11:09 发布

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本文介绍了一个经典的01背包问题案例，通过动态规划算法解决选择最佳魅力饰品组合的问题，在重量限制下最大化饰品的总吸引力。

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 41212		Accepted: 17929

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

Sample Output

01背包水题

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[40010];
int D[40010];
int W[40010];
int main(){
	//freopen("in.txt","r",stdin);
	int n,m;
	while(cin>>n>>m){
		memset(dp,0,sizeof(dp));
		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]);
			}
		}
		//cout<<dp[m]<<endl;
		int ans=-1;
		for(int i=1;i<=m;i++) ans=max(ans,dp[i]);
		cout<<ans<<endl;
	}
}