Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5589 Accepted Submission(s): 3043
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：N个任务M天完成，每个任务花费每天都有一定的收益，问收益最大 分析：分组背包 把N个任务看成N组，其中每组中只能选择一个，也就是每一个任务花费的天数肯定是一个数，然后花费的天数还有费用，分组背包纯裸模板 http://www.cppblog.com/Onway/archive/2010/08/09/122695.html这个讲讲解了第二重和三重循环的设计思路 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 const int MAX = 110; 7 int n,m; 8 int dp[MAX],a[MAX][MAX]; 9 int main() 10 { 11 while(scanf("%d%d", &n, &m) != EOF) 12 { 13 if(n == 0 && m == 0) 14 break; 15 for(int i = 1; i <= n; i++) 16 for(int j = 1; j <= m; j++) 17 scanf("%d", &a[i][j]); 18 memset(dp, 0, sizeof(dp)); 19 for(int i = 1; i <= n; i++) 20 { 21 for(int j = m; j > 0; j--) 22 { 23 for(int k = 1; k <=m; k++) 24 { 25 if(j >= k) 26 dp[j] = max(dp[j], dp[j - k] + a[i][k]); 27 } 28 } 29 } 30 printf("%d\n", dp[m]); 31 } 32 return 0; 33 } View Code
转载于:https://www.cnblogs.com/zhaopAC/p/5047674.html
