【hdu1712】ACboy needs your help——分组背包

最新推荐文章于 2022-08-07 12:11:36 发布

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最新推荐文章于 2022-08-07 12:11:36 发布

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分类专栏：题解心得分组背包文章标签： dp hdu 分组背包

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分组背包

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针对ACboy面临的课程学习时间分配问题，本篇介绍了一种利用动态规划算法进行最大利润优化的方法。通过三维矩阵循环实现物品（课程）在有限天数（背包容量）内的最佳分配。

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题目：

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5819 Accepted Submission(s): 3177

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

Sample Output

Source

HDU 2007-Spring Programming Contest

描述：给定容量为M背包，给出第i个物品消耗为j时可获得的价值，求可获得的最大价值

题解：dp[i][j]表示将前i个物品装入容量为j的背包时的消耗，因为这里的j被赋予了别的含义，所以需要第三重循环来表示物品的消耗。

联系到01背包的空间优化，用一维数组滚动记录即可。注意dp要初始化。

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int maxn = 105;
int a[maxn][maxn], dp[maxn];
int main()
{
	//freopen("input.txt", "r", stdin);
	int N, M;
	while (scanf("%d%d", &N, &M), N || M)
	{
		for (int i = 1; i <= N; i++)
			for (int j = 1; j <= M; j++)
				scanf("%d", &a[i][j]);
		memset(dp, 0, sizeof(dp));
		for (int i = 1; i <= N; i++)
			for (int j = M; j >= 1; j--)
				for (int k = 1; k <= j; k++)
					dp[j] = max(dp[j], dp[j - k] + a[i][k]);
		printf("%d\n", dp[M]);
	}
	return 0;
}