一维的01背包问题

2015年寒假集训第一次周赛的一道题：

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

4 6
1 4
2 6
3 12
2 7

Sample Output

23

 

 

#include
int max(int a,int b)
{
    return a>b?a:b;
}
int w[3405],d[3405],sum[13000];           //（当时做错的原因，就是没有理解透sum这个数组是到底是记录啥的，所以数组开的不够大，导致结果是运行错 误！！）
int main()
{
    int n,m,i,j;
    scanf("%d %d",&n,&m);
    for(i=0; i=w[i]; j--)
            sum[j]=max(sum[j],sum[j-w[i]]+d[i]);            //要充分理解这一行代码的意思（放与不放此时包的最大价值）
    printf("%d\n",sum[m]);
}

1、这是一道最基本的01背包问题，运用一维数组会比二维数组在时间复杂度上减少不少。

2、而sum数组就是用来记录当体积取j时这个包内物品最大的价值。

3、所以sum这个数组要根据所允许的最大体积开内存。