C

最新推荐文章于 2024-11-18 12:43:25 发布

原创最新推荐文章于 2024-11-18 12:43:25 发布 · 223 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#背包问题

背包问题专栏收录该内容

0 篇文章

订阅专栏

本文介绍了一个经典的背包问题求解方法，通过使用动态规划算法，在给定的重量限制内选择一系列物品（在这里为魅力饰物），使得总价值最大化。具体实现包括了如何设置循环来更新最大价值。

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

背包问题
核心是在双循环中比较加或不加第i件物品的价值
要注意c[]的下标是重量，要开的大一点

//背包问题 
#include<stdio.h>
int max(int x,int y)
{
    return x>y?x:y;
}
int main()
{
    int n,m,i,j;
    int w[4000],d[4000];    //w[]存重量，d[]存价值 
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int c[20000]={0};
        //赋零值用于比较，下标是重量
        for(i=1;i<=n;i++)
            scanf("%d%d",&w[i],&d[i]);
        //重点预警 
        for(i=1;i<=n;i++)           //n件物品循环判断 
            for(j=m;j>=w[i];j--)    //循环重量，从0到m，不断比较加/不加第i件物品的价值，c[j]取较大值 
                c[j]=max(c[j],c[j-w[i]]+d[i]);
        printf("%d\n",c[m]);
    }
    return 0;
}