分组背包-ACboy needs your help

最新推荐文章于 2022-08-07 12:11:36 发布

ZXLS-ZMR

最新推荐文章于 2022-08-07 12:11:36 发布

阅读量692

点赞数

CC 4.0 BY-SA版权

分类专栏： C 文章标签：动态规划背包问题

本文链接：https://blog.youkuaiyun.com/chj_zmr/article/details/8785386

C 专栏收录该内容

83 篇文章

订阅专栏

本文探讨了一个关于课程安排的问题，即如何在有限的学习天数内通过合理安排不同课程的学习时间来实现最大化的收益。介绍了输入数据格式，包括课程数量、可用天数及不同天数学习各课程所能获得的收益，并给出了解决方案的代码示例。

Description

ACboy has N courses this term, and he plans to spend at most M days on study.Of course,the profit he will gain from different course depending on the days he spend on it.How to arrange the M days for the N courses to maximize the profit?

input

The input consists of multiple data sets. A data set starts with a line containing two positive integers N and M, N is the number of courses, M is the days ACboy has.
Next follow a matrix A[i][j], (1<=i<=N<=100,1<=j<=M<=100).A[i][j] indicates if ACboy spend j days on ith course he will get profit of value A[i][j].
N = 0 and M = 0 ends the input.

output

For each data set, your program should output a line which contains the number of the max profit ACboy will gain.

same input

2 2 1 2 1 3 2 2 2 1 2 3 3 2 1 3 2 1 0 0

output

3 4 6

#include<string.h>
#include<stdio.h>
int main()
{  
  int a[105][105],f[105];
  int i,j,n,m,v;
  while(scanf("%d%d",&n,&m),n|m)
   {
    for(i=1;i<=n;i++)
     for(j=1;j<=m;j++)
      scanf("%d",&a[i][j]);
     memset(f,0,sizeof(f));
     for(i=1;i<=n;i++)                     //从n个课程里选择。
      for(j=m;j>=0;j--)                    //剩下的天数
       for(v=1;v<=j;v++)                   //从每组里选择。
        f[j]=f[j]>f[j-v]+a[i][v]?f[j]:f[j-v]+a[i][v];
     printf("%d\n", f[m]);
         }
   return 0;
}