ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3282    Accepted Submission(s): 1703

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

分組揹包！....

 1 /*o1揹包@龔細軍*/
 2 /*維度為2的01揹包*/
 3 #include<stdio.h>
 4 #include<string.h>
 5 #define maxn 102
 6 int dp[maxn];
 7 int aa[maxn][maxn];
 8 
 9 int max(int a,int b)
10 {
11     return a>b?a:b;
12 }
13 int main()
14 {
15     int m,n,i,j,k;
16     while(scanf("%d%d",&n,&m),m+n)
17     {
18         memset(dp,0,sizeof(dp));
19         for(i=1;i<=n;i++)        // class
20             for(j=1;j<=m;j++)    //day
21                 scanf("%d",&aa[i][j]);
22             //對每一門課程進行揹包施放
23             for(i=1;i<=n;i++)
24             {
25                for(j=m;j>=0;j--)  //代表的是揹包的容量
26                {
27                    for(k=1;k<=j;k++)  //在這個容量內選擇一個房間去，和之前放進去的比較！
28                        dp[j]=max(dp[j],dp[j-k]+aa[i][k]);
29                }
30             }
31          printf("%dn",dp[m]);
32     }
33     return 0;
34 }