ACboy needs your help（HDU-1712）

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

 

題意：學生有n門課，有m天時間看，每門課看幾天產生多大的價值，求最大價值；

分組揹包：
這個問題變成了每組物品有若干種策略：是選擇本組的某一件，還是一件都不選。也就是說設f[k][v]表示前k組物品花費費用v能取得的最大權值，則有：
f[k][v]=max{f[k-1][v],f[k-1][v-c[i]]+w[i]|物品i屬於組k}

#include<iostream>
#include<cstring>
using namespace std;
int n,m;
int a[105][105],dp[105];
int main()
{
    while(cin>>n>>m)
    {
        if(m==0&&n==0) break;
        memset(dp,0,sizeof(dp));
        memset(a,0,sizeof(a));
        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<=j; k++)
                {
                    dp[j]=max(dp[j],dp[j-k]+a[i][k]);
                }
            }
        }
        cout<<dp[m]<<endl;
    }
    return 0;
}