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): 6953 Accepted Submission(s): 3853Problem 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 0Sample Output
3
4
6Source
HDU 2007-Spring Programming Contest
题意:
n门课程m天时间,每天最多选一门课程,A [i] [j]表示花费j天,获得价值A [i] [j]的价值,如何安排使得价值最大。
题解:分组01背包,分成n组,容量为m,价值为a[i][j],每组选取一门课程花费k天,获得最大价值。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 105;
int n,m;
int a[maxn][maxn];
int dp[maxn];
int main()
{
while(cin>>n>>m&&(m+n))
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
cin>>a[i][j];
}
}
for(int i=1;i<=n;i++)//n组
{
for(int j=m;j>=1;j--)//天数从小到大
{
for(int k=j;k>=1;k--)//01
{
dp[j] = max(dp[j],dp[j-k]+a[i][k]);
}
}
}
cout<<dp[m]<<endl;
memset(dp,0,sizeof(dp));
}
return 0;
}