HDU

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7609 Accepted Submission(s): 4211

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
题意：ACboy 这学期有N个课程可读，但他只有M天的时间去学习，然后给你一个nN∗M的矩阵a，第i行j列表示第i个课程需要花费j天完成，获得a[i][]j]的收益，问你这M天你最多可以获得多少收益

分析： 这道题是比较裸的分组背包问题，具体的算法在我博客里有，简单说下，就是对N个中的每一个课程都有两种选择要么学，要么不学，如果不学的话很好处理直接等于前一个状态即可，如果学的话就保存上，具体的状态转移方程为：dp[j][k]=max(dp[j−1][k],dp[j−1][k−w[i]]+v[i])代表第j个课程要么选，要么不选，这里的话用一个一位的数组直接滚动下，就可以省掉一维，类似于01背包

参考代码

#include<bits/stdc++.h>using namespace std;const int N = 1e4 + 10;int a[105][105],dp[N];int main(){    int n,m;    while(cin>>n>>m,n+m){        for(int i = 1;i <= n;i++)            for(int j = 1;j <= m;j++)                cin>>a[i][j];        memset(dp,0,sizeof(dp));        for(int i = 1;i <= n;i++)            for(int k = m;k >= 0;k--)                for(int j = 1;j <= k;j++) //这里j就是相当于”体积“                    dp[k] = max(dp[k],dp[k-j]+a[i][j]);        cout<<dp[m]<<endl;    }    return 0;}

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