第六讲 分组的背包问题 HDU 1712ACboy needs your help

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5618    Accepted Submission(s): 3063

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

 

问题

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属于组k}

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]}

代码如下：

#include<iostream>//c++#include<cmath>//数学公式#include<cstdlib>//malloc#include<cstring>#include<string>#include<cstdio>//输入输出#include<algorithm>//快排#include<queue>//队列#include<functional>//优先队列#include<stack>//栈#include<vector>//容器#include<map>//地图  if continuetypedef long long ll;const  int N=105;using namespace std;int dp[N],value[N][N];int main(){//    freopen("C:\\Users\\ch\\Desktop\\1.txt","r",stdin);//freopen("C:\\Users\\lenovo\\Desktop\\2.txt","w",stdout);int i,j,k;int v,n,m;while(cin>>m>>n,n|m)    {        for(i=0;i<m;i++)        for(j=1;j<=n;j++)    scanf("%d",&value[i][j]);        memset(dp,0,sizeof(dp));        for(int i=0;i<m;i++)        for(int j=n;j>=1;j--)        for(int k=1;k<=j;k++)            dp[j]=max(dp[j],dp[j-k]+value[i][k]);        cout<<dp[n]<<endl;    }    return 0;}