第四讲 混合三种背包问题 HDU 3535 AreYouBusy

AreYouBusy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3643    Accepted Submission(s): 1428

Problem Description

Happy New Term!
As having become a junior, xiaoA recognizes that there is not much time for her to AC problems, because there are some other things for her to do, which makes her nearly mad.
What's more, her boss tells her that for some sets of duties, she must choose at least one job to do, but for some sets of things, she can only choose at most one to do, which is meaningless to the boss. And for others, she can do of her will. We just define the things that she can choose as "jobs". A job takes time , and gives xiaoA some points of happiness (which means that she is always willing to do the jobs).So can you choose the best sets of them to give her the maximum points of happiness and also to be a good junior(which means that she should follow the boss's advice)?

 

Input

There are several test cases, each test case begins with two integers n and T (0<=n,T<=100) , n sets of jobs for you to choose and T minutes for her to do them. Follows are n sets of description, each of which starts with two integers m and s (0<m<=100), there are m jobs in this set , and the set type is s, (0 stands for the sets that should choose at least 1 job to do, 1 for the sets that should choose at most 1 , and 2 for the one you can choose freely).then m pairs of integers ci,gi follows (0<=ci,gi<=100), means the ith job cost ci minutes to finish and gi points of happiness can be gained by finishing it. One job can be done only once.

 

Output

One line for each test case contains the maximum points of happiness we can choose from all jobs .if she can’t finish what her boss want, just output -1 .

 

Sample Input

   
   
   
   

    
    
    
    3 3
2 1
2 5
3 8
2 0
1 0
2 1
3 2
4 3
2 1
1 1

3 4
2 1
2 5
3 8
2 0
1 1
2 8
3 2
4 4
2 1
1 1

1 1
1 0
2 1

5 3
2 0
1 0
2 1
2 0
2 2
1 1
2 0
3 2
2 1
2 1
1 5
2 8
3 2
3 8
4 9
5 10
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    5
13
-1
-1
   
   
   
   

 

问题：

如果将P01、P02、P03混合起来。也就是说，有的物品只可以取一次（01背包），有的物品可以取无限次（完全背包），有的物品可以取的次数有一个上限（多重背包）。应该怎么求解呢？这就是一个混合背包。

具体问题具体分析。

清晰思路：

for i=1..N

    if 第i件物品属于01背包

        ZeroOnePack(c[i],w[i])

    else if 第i件物品属于完全背包

        CompletePack(c[i],w[i])

    else if 第i件物品属于多重背包

        MultiplePack(c[i],w[i],n[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 continue
 typedef long long ll;
 const  int N=105;
 using namespace std;
 int dp[N],d[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,g;
     while(~scanf("%d%d",&n,&v))
     {
         memset(dp,0,sizeof(dp));
         for(k=0;k<n;k++)
         {
             cin>>m>>g;
             if(g==0)
             {
                 memcpy(d,dp,sizeof(dp));
                 memset(dp,-66,sizeof(dp));
                 for(i=0;i<m;i++)
                 {
                     int a,b;
                     cin>>a>>b;
                     for(j=v;j>=a;j--)
                         dp[j]=max(dp[j],max(dp[j-a]+b,d[j-a]+b));
                 }
                 continue;
             }
             if(g==1)
             {
                 memcpy(d,dp,sizeof(dp));
                 for(i=0;i<m;i++)
                 {
                     int a,b;
                     cin>>a>>b;
                     for(j=v;j>=a;j--)
                         dp[j]=max(dp[j],d[j-a]+b);
                 }
                 continue;
             }
             if(g==2)
             {
                 for(i=0;i<m;i++)
                 {
                     int a,b;
                     cin>>a>>b;
                     for(j=v;j>=a;j--)
                         dp[j]=max(dp[j],dp[j-a]+b);
                 }
                 continue;
             }
         }
         dp[v]=max(dp[v],-1);
         cout<<dp[v]<<endl;
     }
     return 0;
 }