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
分组背包!....
1 /*o1背包@龚细军*/
2 /*维度为2的01背包*/
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 //对每一门课程进行背包施放
23 for(i=1;i<=n;i++)
24 {
25 for(j=m;j>=0;j--) //代表的是背包的容量
26 {
27 for(k=1;k<=j;k++) //在这个容量内选择一个房间去,和之前放进去的比较!
28 dp[j]=max(dp[j],dp[j-k]+aa[i][k]);
29 }
30 }
31 printf("%d\n",dp[m]);
32 }
33 return 0;
34 }