hdu 1712 ACboy needs your help（分组背包入门）

最新推荐文章于 2022-08-07 12:11:36 发布

AC_way

最新推荐文章于 2022-08-07 12:11:36 发布

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分类专栏： ACM-分组背包文章标签： algorithm

本文链接：https://blog.youkuaiyun.com/caduca/article/details/39187967

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本文介绍了一道关于课程安排的问题，ACboy需要在有限的时间内安排课程以获得最大收益。文章提供了一个通过分组背包算法解决该问题的C++代码示例。

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ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4185 Accepted Submission(s): 2220

Problem 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.

Sample Input

Sample Output

3
4
6

分组背包入门题

代码：

//15ms
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn=100+10;
int a[maxn][maxn];
int dp[maxn];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0)
        break;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            scanf("%d",&a[i][j]);
        }
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)//根据不同课程分组
        {
            for(int v=m;v>=1;v--)//枚举不同容量
            {
                for(int j=1;j<=v;j++)//不同时间的课程
                {
                    dp[v]=max(dp[v],dp[v-j]+a[i][j]);
                }
            }
        }
        printf("%d\n",dp[m]);
    }
    return 0;
}