ACboy needs your help
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5964 Accepted Submission(s): 3251
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.
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
题目大意：ACboy要开始选课了，上一门课能够获得的收益和他上这门课的时间是有关的，然后给你若干门课，让你帮他进行选课，
每一门课自然是只能选择一个课程时长，问你如何选择，才能使ACboy获得的受益最大。
思路分析：这是自己做的第一道分组背包的题目，抽象一下，背包容量相当于ACboy拥有的时间，然后给你若干组物品（课程），每一组
物品里面有着若干个物品（课程时间），一组里面的物品只能选择一个，这么一抽象，标准的分组背包问题，相当于若干组物品的01背包
问题，但是要注意，v的循环一定要放在组内物品循环之外，这样才能保证一个组的物品只用了一个，另外，内层循环要有一个判断，判断是
否满足v-cost>=0，否则会出现数组越界，报错。
代码：#include <iostream>
#include <stack>
#include <cstdio>
#include <cstring>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn=105;
int a[maxn][maxn];
int f[maxn];
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)&&(n||m))
{
memset(f,0,sizeof(f));
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++)
{
for(int j=m;j>=0;j--)
{
for(int k=1;k<=m;k++)
{
if(j-k>=0)
f[j]=max(f[j],f[j-k]+a[i][k]);
}
}
}
cout<<f[m]<<endl;
}
}
#include <stack>
#include <cstdio>
#include <cstring>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn=105;
int a[maxn][maxn];
int f[maxn];
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)&&(n||m))
{
memset(f,0,sizeof(f));
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++)
{
for(int j=m;j>=0;j--)
{
for(int k=1;k<=m;k++)
{
if(j-k>=0)
f[j]=max(f[j],f[j-k]+a[i][k]);
}
}
}
cout<<f[m]<<endl;
}
}