HDU-1712 ACboy needs your help (分组背包 入门题)

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7709    Accepted Submission(s): 4255

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 21 21 32 22 12 12 33 2 13 2 10 0

 

Sample Output

346

 

#include <stdio.h>#include <bits/stdc++.h>using namespace std;int a[101][101], dp[101];int main(){int n, m;while(scanf("%d %d", &n, &m) != EOF){if(n == 0 && m == 0) return 0;memset(dp, 0, sizeof(dp));memset(a, 0, sizeof(a));for(int i = 1; i <= n; ++i){for(int j = 1; j <= m; ++j){scanf("%d", &a[i][j]);}}for(int i = 1; i <= n; ++i){for(int j = m; j >= 0; --j){for(int k = 1; k <= j; ++k){dp[j] = max(dp[j], dp[j - k] + a[i][k]);}}}printf("%d\n", dp[m]);}}/*题意：100门课，每门课上不同天数收益不同，一共100天，如何安排可以使收益最大。思路：分组背包裸题。只需要在更新的时候保证每一组中的物品只有一个更新当前状态即可。该题是同一门课不同天数的收益为一组，每一门课只能选一次上课的天数。*/