经典背包问题 POJ 3624

最新推荐文章于 2017-09-09 21:59:05 发布

最新推荐文章于 2017-09-09 21:59:05 发布 · 93 阅读

本文介绍了一种解决魅力手链问题的动态规划算法，旨在帮助Bessie在给定重量限制下选择最佳魅力饰品组合，以实现最高的魅力值总和。通过使用滚动数组优化内存使用，确保了算法的有效性和实用性。

Charm Bracelet

Time Limit:1000MS		Memory Limit:65536K
Total Submissions:6996		Accepted:3174

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 theN(1 ≤N≤ 3,402) available charms. Each charmiin the supplied list has a weightW_i(1 ≤W_i≤ 400), a 'desirability' factorD_i(1 ≤D_i≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more thanM(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:NandM
* Lines 2..N+1: Linei+1 describes charmiwith two space-separated integers:W_iandD_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

用滚动数组写的，否则超内存

#include<stdio.h>
int opt[3][12881];
int value[3450];
int weight[3450];


int main()
{
    int n,v;
    int i,j;
    freopen("input","r",stdin);
    scanf("%d %d",&n,&v);
    for(i=0;i<n;i++)
        scanf("%d %d",&weight[i],&value[i]);
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=v;j++)
        {
            if(j>=weight[i-1])
                opt[i%3][j]=(opt[(i-1)%3][j]>opt[(i-1)%3][j-weight[i-1]]+value[i-1]?opt[(i-1)%3][j]:opt[(i-1)%3][j-weight[i-1]]+value[i-1]);
            else
                opt[i%3][j]=opt[(i-1)%3][j];
        }
    }
    printf("%d",opt[n%3][v]);
    return 0;
}