题目信息:
ACboy needs your help
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5146 Accepted Submission(s): 2780
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:
一个学生用M天的时间复习N门课程,每门课程花费不同的天数,有不同的收获。问如何安排这M天,使得收获最大。
解题思路:
分组背包即物品分为多组,每一组的各个物品最多只能取一个。解题方法即在循环最里面再嵌套一个循环,查找每组中价值最大的物品。
代码部分:
#include <stdio.h>
#include <string.h>
int main()
{
int N,M,i,j,k;
int dp[105],value[105];
while(scanf("%d %d",&N,&M)&&M||N)
{
memset(dp,0,sizeof(dp));
memset(value,0,sizeof(value));
for(i=1;i<=N;i++)
{
for(j=1;j<=M;j++)
{
scanf("%d",&value[j]);
}
for(j=M;j>0;j--)
{
for(k=j;k>0;k--)
dp[j]=dp[j]>dp[j-k]+value[k]?dp[j]:dp[j-k]+value[k];
}
}
printf("%d\n",dp[M]);
}
return 0;
}