ACboy needs your help（分组背包）-优快云博客

本文链接：https://blog.youkuaiyun.com/iteye_6881/article/details/82543956

本文探讨了一种课程选择与时间分配的问题，即如何在有限的时间内通过选择不同课程的不同学习天数来最大化收益。使用分组背包算法解决该问题，并提供了一个具体的实现示例。

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ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3109 Accepted Submission(s): 1619

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

题意：

输入为多项，给出N门课程（1到100）和M天时间（1到100），随后给出一个N*M的矩阵，A[ i ][ j ]代表花费 j 天可以获得A[ i ][ j ]的分数。但是每门课程只能选一个时间，就是说每行只能选一个数。求在不超过有限时间内的最大成绩。

思路：

分组背包。

AC：

#include<stdio.h>
#include<string.h>
int c[105][105];
int dp[105];
int main()
{
	int n,m;
	while(scanf("%d%d",&n,&m)!=EOF&&(n+m))
	{
	  memset(dp,0,sizeof(dp));
	  memset(c,0,sizeof(c));
	  for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		  scanf("%d",&c[i][j]);
	  for(int i=1;i<=n;i++)
		for(int j=m;j>=0;j--)  
		   for(int k=1;k<=j;k++)  //一组中的每个都要进行一次01背包
			  dp[j]=(dp[j]>dp[j-k]+c[i][k]?dp[j]:dp[j-k]+c[i][k]);
	  printf("%d\n",dp[m]);
	}
	return 0;
}