HDU 1712 ACboy needs your help

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文章详细阐述了如何在有限的时间内，通过合理安排课程学习，最大化学业收益的策略。通过解决一个具体的数学问题，读者可以了解到如何运用分组背包算法来优化时间分配，从而在学习上获得更高的效率和成果。

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ACboy needs your help

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include <queue>

using namespace std;
#define maxn 105
#define inf 0x7ffffff
pair <int ,int > p;
int arr[maxn][maxn];
int dp[maxn];
int n,m;
//int cmpD(Node a,Node b){
//    return a.h > b.h;
//}
//int cmpI(Node a,Node b){
//    return a.h < b.h;
//}
int main()
{
    while(scanf("%d%d",&n,&m) != EOF && m+n){
        if(m == 0 && n == 0){
            break;
        }
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m;j++){
                scanf("%d",&arr[i][j]);
            }
        }
        memset(dp,0,sizeof(dp));
            for(int i = 1; i <= n; i++){
                for(int j = m; j >= 0; j--){
                    for(int k = 1; k <= j; k++){
                        dp[j] = max(dp[j],dp[j - k] + arr[i][k]);
                    }
                }
            }
        printf("%d\n",dp[m]);
    }
    return 0;
}