hdu 1712 ACboy needs your help 分组背包

题目 ：

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

多组背包 ：每组中的物品至多选一个

二维代码 ：（ 用原来的那个数组直接更新 ）

#include<iostream>

using namespace std;

int main( ){

    int n,m,dp[105][105]={0},Max;
    while(1) {

        cin >>n >>m;
        if( n==0 && m==0 ) break;
        for( int i=1;i<=n;i++)
            for( int j=1;j<=m;j++)
                cin >>dp[i][j];

        for( int i=2;i<=n;i++) {
            for( int j=m;j>=1;j--){
                Max=dp[i][j];
                //  对 第i行第j行 位置更新 
                //  因为dp[i-1][j-k] 是 第i-1行第j-k行 位置 的最优解 ,所以对于 一个dp[i][k] 应该对应的 + dp[i-1][j-k]，得一个较大值。
                for( int k=1;k<j;k++)   
                    Max=max( Max, dp[i-1][j-k]+dp[i][k] );
                dp[i][j]=max( Max, dp[i-1][j] ) ;
            }
        }
        Max=-1;

        for( int i=1;i<=m;i++)
            Max=max( Max, dp[n][i] );
        cout <<Max <<endl;
    }
}

一维数组 ：

#include<iostream>
#include<cstring>

using namespace std;

int main( ){

    int n,m,V[105][105]={0},dp[105];
    while(1) {

        cin >>n >>m;
        if( n==0 && m==0 ) break;
        memset (dp,0,sizeof(dp));

        for( int i=1;i<=n;i++)
            for( int j=1;j<=m;j++)
                cin >>V[i][j];

        for( int i=1;i<=n;i++)
            for( int j=m;j>=1;j--)
                for( int k=0;k<=j;k++)
                    dp[j]=max( dp[j], dp[j-k]+V[i][k] );
        for( int i=1;i<=m;i++)
           dp[0]=max( dp[0],dp[i] );

        cout <<dp[0] <<endl;
    }
}