HDU 1712 ACboy needs your help (分组背包 每组至多选一个)

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6444    Accepted Submission(s): 3564

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

 

Source

HDU 2007-Spring Programming Contest

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1712

题目大意：一个人有n门课，最多学m天，对第i门课学j天的成绩为a[i][j]，求如何安排m天可以使成绩最高

题目分析：分组背包问题，每门课是一组，设dp[i]表示花了i天能得到的最高成绩，想要保证每组至多选一个，只需将对每组的枚举放在最内层循环

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int const MAX = 105;int n, m;int dp[MAX], a[MAX][MAX];int main() {    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 >= 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]);    }}