hdu1712分组背包

B - ACboy needs your help
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
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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：学生A,有n科考试，m天复习时间，然后有一个n*m的矩阵，a[i][j]表示第i科考试用j天复习得到的效益值，求最大效益值
，显然每一科只能选择确定天数去复习；
这便是分组背包了：

   有N件物品和一个容量为V的背包。第i件物品的费用是c[i]，价值是w[i]。这些物品被划分为若干组，每组中的物品互相冲突，最多选一件。求解将哪些物品装入背包可使这些物品的费用总和不超过背包容量，且价值总和最大。

算法

这个问题变成了每组物品有若干种策略：是选择本组的某一件，还是一件都不选。也就是说设f[k][v]表示前k组物品花费费用v能取得的最大权值，则有：

f[k][v]=max{f[k-1][v],f[k-1][v-c[i]]+w[i]|物品i属于第k组}

使用一维数组的伪代码如下：

for 所有的组k

    for
v=V..0

        for 所有的i属于组k

            f[v]=max{f[v],f[v-c[i]]+w[i]}

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int max(int a,int b)
{
if(a>b)
return a;
return b;
}
const int maxn=10100,inf=0x3f3f3f3f;
int dp[maxn],a[105][105];
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
if(n==0&&m==0)break;
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>=1;j--)
{
for(int k=1;k<=m;k++)
{
if(j-k>=0)
dp[j]=max(dp[
}
}j],dp[j-k]+a[i][k]);
}
printf("%d\n",dp[m]);
}

return 0;
}