ACboy needs your help

最新推荐文章于 2020-08-26 17:18:22 发布

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#动态规划

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本文探讨了一种算法，用于解决ACboy如何在有限的学习天数内，通过合理安排不同课程的学习时间，以达到利润最大化的策略。输入包括课程数量、可用学习天数及各课程在不同天数学习下所能获得的利润矩阵。算法通过动态规划，计算并输出最大可能的总利润。

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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
这个题有点难想啊，是我太笨了。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<math.h>
using namespace std;
int a[105][105];
int dp[105];
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)){
        if(n==0&&m==0){return 0;}
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                scanf("%d",&a[i][j]);
            }
        }
        int mmax=0;
        for(int i=1;i<=n;i++){
            for(int j=m;j>=1;j--){
                for(int k=1;k<=j;k++){
                    dp[j]=max(dp[j],dp[j-k]+a[i][k]);
                    //mmax=max(mmax,dp[j]);
                    //cout<<dp[j]<<" ";
                }
            }
            //cout<<endl;
        }
        printf("%d\n",dp[m]);
    }
}