HDU_1712_背包问题

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本文介绍了一个课程学习计划的问题，通过动态规划算法实现最优方案选择，旨在最大化学习收益的同时不超出限定的学习天数。

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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

Sample Output

 3
4
6

题意就是告诉你每门课程话费一定天数，所能获得的知识。

让你列一个计划，让你所能获得的知识最大

dp[i][j]的意义是学习前i们课程话费j天所能获得的最大知识

我们首先应该明确状态转移方程

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

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define MAX_N 105
#define MAX_M 105
int n;//课程数目
int m;//总天数
//dp[i+1][j]表示学习前i门课程消耗j天所获得最大价值
/**
    //状态转移方程
    dp[i+1][j]=max(dp[i][j],dp[i][j-k]+A[i][k]|k<=j)
*/
int dp[MAX_N][MAX_M];

//A[i][j]代表花j天在第i门课上所获得的价值
int A[MAX_N][MAX_M];
int main()
{
    //freopen("背包D.txt","r",stdin);
    while(~scanf("%d %d",&n,&m))
    {
        if(n==m&&m==0)
        {
            break;
        }
        for(int i=0;i<n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                scanf("%d",&A[i][j]);
            }
        }
        for(int i=0;i<n;i++)
        {
            A[i][0]=0;
        }
        void solve();
        solve();
    }

}
void solve()
{
    memset(dp,0,sizeof(dp));
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<=m;j++)
        {

<span style="white-space:pre">	</span>    //对所能学习的天数进行枚举
            for(int k=0;k<=j;k++)
            {
                //dp[i][j]代表学习前i门课程所占j天所获得的最大价值
                dp[i+1][j]=max(dp[i+1][j],dp[i][j-k]+A[i][k]);
            }
        }
    }
    printf("%d\n",dp[n][m]);
}