0
点赞
收藏
分享

微信扫一扫

Charm Bracelet 3624 (01背包)

juneyale 2023-04-20 阅读 54


Charm Bracelet


Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u


Submit  Status


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




#include<stdio.h>
#include<string.h>
#define MAX(a,b) a>b?a:b
int w[35000],v[35000],sum[35000];
int main()
{
	int n,m,i,j;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(sum,0,sizeof(sum));
		for(i=1;i<=n;i++)
			scanf("%d%d",&w[i],&v[i]);
		for(i=1;i<=n;i++)
		{
			for(j=m;j>=w[i];j--)
			{
				sum[j]=MAX(sum[j],sum[j-w[i]]+v[i]);
			}
		}
		printf("%d\n",sum[m]);
	}
	return 0;
}



举报

相关推荐

0 条评论