ACboy needs your help（分组背包）

最新推荐文章于 2020-08-26 17:18:22 发布

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#分组背包

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本文介绍了一个课程安排优化问题，通过动态规划求解如何在有限的学习天数内安排多个课程以获得最大收益。输入包括课程数量及每天的学习效益，输出则是最大可能的总效益。

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

伪代码：for 1...i组
for V....0
for 所有的k属于i组
dp[j]=max(dp[j],dp[j-v[i]]+w[i]);

#include<stdio.h>
#include<string.h>
#define Max 105
int w[Max][Max],dp[Max];
int m,n;
int main()
{
	//freopen("b.txt","r",stdin);
	int i,j,k;
	while(scanf("%d %d",&m,&n)==2)
	{
		if(m+n==0) break;
		for(i=1;i<=m;i++)
			for(j=1;j<=n;j++)
				scanf("%d",&w[i][j]);
			memset(dp,0,sizeof(dp));
			for(i=1;i<=m;i++)
				for(j=n;j>=0;j--)
				{
					for(k=1;k<=n;k++)
						if(j>=k) dp[j]=dp[j]>dp[j-k]+w[i][k]?dp[j]:dp[j-k]+w[i][k];
				}
				printf("%d\n",dp[n]);
	}
	return 0;
}