[HDU#1712]ACboy needs your help

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6038    Accepted Submission(s): 3297

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

这是一个分组背包问题,使用DP算法即可实现求解。

#include<iostream>
#define INF 0x7fffffff
#define max(a,b) (a>b?a:b)
using namespace std;
int main(void)
{
#ifndef ONLINE_JUDGE 
 freopen("in.txt", "r", stdin);
#endif
 int n, m;
 while (cin >> n >> m)
 {
  if (n == 0 && m == 0) break;
  int t[101][101] = { 0 };//j days on ith     profit
  int dp[101][101]= { 0 };
  for (int i = 1; i <= n; i++)
   for (int j = 1; j <= m; j++)
   {
    cin >> t[i][j];
   }
  for (int i = 1; i <= n; i++)
  {
   for (int j = 1; j <= m; j++)
   {
    for (int k = 0; k <= j; k++)
    {
     dp[i][j] = max(dp[i][j], dp[i - 1][j - k] + t[i][k]);
    }
   }
  }
  cout << dp[n][m] << endl;
 }
 return 0;
}

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