HDU 1712 ACboy needs your help

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#动态规划 #背包 #HDU #NOIP

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本文探讨了一种典型的分组背包问题，即如何在有限的时间内，通过合理分配时间到不同的课程上，以最大化收益。文章提供了详细的算法实现，包括使用三维数组进行状态转移的过程。

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题目链接：传送门
题面：

ACboy needs your help

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

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

Sample Output

3
4
6

题意：

你有 $N$ 天要学 $M$ 门课， $a [i] [j]$ 表示第 $i$ 门课花了 $j$ 天你可以获得的价值，求如何选择课程能使得你获得的价值最大。

分组背包
一组物品就是每一门课的不同价值
只能取一次
先枚举组数
再枚举体积
再枚举每组中的物品

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <complex>
#include <algorithm>
#include <climits>
#include <queue>
#include <map>
#include <vector>
#include <iomanip>
#define A 10010
#define B 110
#define ll long long

using namespace std;
int f[A], n, m, a[B][B];

int main() {
    while (scanf("%d%d", &n, &m)) {
        if (!n and !m) return 0;
        memset(f, 0, sizeof f);
        for (int i = 1; i <= n; i++)
          for (int j = 1; j <= m; j++)
            scanf("%d", &a[i][j]);
        for (int i = 1; i <= n; i++)
          for (int j = m; j >= 0; j--)
            for (int k = 1; k <= m; k++)
              f[j] = max(f[j], f[j - k] + a[i][k]);
        printf("%d\n", f[m]);
    }
}