hdu1712ACboy needs your help( dp )

题目大意就是给出 N门课， 然后给出每门课花m天的收益，问在天数一定的情况下最大收益

ans[x][r] 表示前x节课在r天的情况下的最佳情况

for ( int i= 0, r)

ans[x][r] = max( ans[x][r] , ans[x+1][r-i] + cl[x][i])

用了for和dfs两种方法写

题目：

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3368    Accepted Submission(s): 1742

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

 

Source

HDU 2007-Spring Programming Contest

 

Recommend

lcy

 

代码：

//Sat Jan 11 18:55:27 2014
//Author:Minshik
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <memory.h>
#include <cctype>
#include <cmath>
#include <list>
using namespace std;

int TIAN,KE;
int cl[105][105];
int ans[105][110];

#define REP(i,N) for(int i=0;i<N;i++)

int dfs(int x,int r)
{
    if(ans[x][r]) return ans[x][r];

    if(x==KE)return 0;

    int &d = ans[x][r];

    d = dfs(x+1,r);

    for(int i=1;i<=r;i++)
    {
        d = max(d,dfs(x+1,r-i)+cl[x][i-1]);
    }
    return d;

}
int main()
{

    while(cin>>KE>>TIAN && KE+TIAN!=0)
    {
        REP(i,KE)REP(j,TIAN)
                cin>>cl[i][j];



        for(int i=1;i<=TIAN;i++)
        {
            for(int j=1;j<=KE;j++)
            {
                 ans[j][i] = ans[j-1][i];
                for(int k=1;k<=i;k++)
                {
                     ans[j][i] = max (ans[j][i], ans[j-1][i-k] +cl[j-1][k-1] );
                }

            }
        }




        cout<<ans[KE][TIAN]<<endl;


    }


    return 0;
}