Charm Bracelet POJ – 3624 (01背包问题)

作者: zqqian 分类: ACM,算法 发布时间: 2017-04-26 19:19

Version:1.0 StartHTML:000000193 EndHTML:000002773 StartFragment:000000845 EndFragment:000002715 StartSelection:000000845 EndSelection:000002715 SourceURL:http://poj.org/problem?id=36243624 — Charm Bracelet

Charm Bracelet
Time Limit: 1000MSMemory Limit: 65536K
Total Submissions: 38622Accepted: 16751

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 鈮?N 鈮?3,402) available charms. Each charm i in the supplied list has a weight Wi (1 鈮?Wi 鈮?400), a ‘desirability’ factor Di (1 鈮?Di 鈮?100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 鈮?M 鈮?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

题目很简单,就是01背包的基础题目。用动态规划就可以做。

#include<iostream>
using namespace std;
struct charm{
	int w;
	int v;
}a[3505];
int max2(int a,int b){
	if(a>b)
	return a;
	return b;
}
int main(){
 int n,m;
 cin>>n>>m;
 
 for(int i=1;i<=n;i++){
 	cin>>a[i].w>>a[i].v;
 }	
 int f[12890]={0};
 	
 		
	for(int i=1;i<=n;i++){
		
		for(int j=m;j>=a[i].w;j--){
			f[j]=max2(f[j],f[j-a[i].w]+a[i].v);
			
	
		}
	}
cout<<f[m]<<endl;
	return 0;
} 

 

 

 

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