ACboy needs your help(简单DP)

 

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

 

 

题意:第一行两个整数 n,m ,代表有 n 个课程，m 天学习，然后 n 行，每行 m 个数，i 行 j 列代表第 i 课程花费 j 天可以获得的价值。问 m 天可以获得得最大价值

dp[i][j] 代表 i 种书，花费 j 天可以获得的最大价值

dp[i][j] = dp[i-1][j] + max (A[i][k])    0<=k<=j 

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
using namespace std;
#define MX 105
int n,m;
int dp[MX][MX];
int A[MX][MX];
int main()
{
    while (scanf("%d%d",&n,&m)&&(n||m))
    {
        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++) //   i 本书
        {
            for (int j=1;j<=m;j++) // j 天
            {
                for (int k=0;k<=j;k++)
                {
                    dp[i][j]=max(A[i][k]+dp[i-1][j-k],dp[i][j]);
                }
            }
        }
        printf("%d\n",dp[n][m]);
    }
    return 0;
}