HDU-3535-AreYouBusy（分组背包）

阳光穿透心脏的1/2处 2022-05-26 12:50 105阅读 0赞

# AreYouBusy #

#### 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

题目描述：有n个组的工作，有t的时间去做

对于组有三种类型

2随意，拿或者不拿，选其中多少都可以

1最多选一个

0至少选一个

每组每件事有花费的时间和收益

取尽量高的收益

**解题思路：**

经典混合背包

题目给了很多类别的物品。用数组dp\[i\]\[j\],表示第i组，时间为j时的快乐值。每得到一组工作就进行一次DP，所以dp\[i\]为第i组的结果。

**第一类（01背包坑点多）**，至少选一项，即必须要选，那么在开始时，对于这一组的dp的初值，应该全部赋为负无穷，这样才能保证不会出现都不选的情况。

状态转移方程：dp\[i\]\[k\]=max(max(dp\[i\]\[k-c\[j\]\]+g\[j\],dp\[i\]\[k\]),dp\[i-1\]\[k-c\[j\]\]+g\[j\]);

dp\[i\]\[j\]: 是不选择当前工作；

dp\[i-1\]\[k-c\[j\]\]+g\[j\]：第一次在本组中选物品，由于开始将该组dp赋为了负无穷，所以第一次取时，必须由上一组的结果推知，这样才能保证得到全局最优解；

dp\[i\]\[k-c\[j\]\]+g\[j\]：表示选择当前工作，并且不是第一次取;

**第二类（分组背包）**，最多选一项，即要么不选，一旦选，只能是第一次选。

状态转移方程：dp\[i\]\[k\]=max(dp\[i-1\]\[k-c\[j\]\]+g\[j\],dp\[i\]\[k\]);

由于要保证得到全局最优解，所以在该组DP开始以前，应该将上一组的DP结果先复制到这一组的dp\[i\]数组里，因为当前组的数据是在上一组数据的基础上进行更新的。

**第三类（01背包）**，任意选，即不论选不选，选几个都可以。

状态转移方程为：dp\[i\]\[k\]=max(dp\[i\]\[k-c\[j\]\]+g\[j\],dp\[i\]\[k\]);

同样要保证为得到全局最优解，先复制上一组解，数据在当前组更新。

#include<bits/stdc++.h>
    using namespace std;
    #define INF 0x3f3f3f3f
    int dp[105][105];
    int main()
    {
        int n,t,m,s;
        int c[105],g[105];
        while(~scanf("%d%d",&n,&t))
        {
            memset(dp,0,sizeof(dp));
            for(int i=1;i<=n;i++)
            {
                scanf("%d%d",&m,&s);
                for(int j=1;j<=m;j++)
                {
                    scanf("%d%d",&c[j],&g[j]);
                }
                if(s==0)
                {
                    for(int j=0;j<=t;j++)
                    {
                        dp[i][j]=-INF;
                    }
                    for(int j=1;j<=m;j++)
                    for(int k=t;k>=c[j];k--)
                    {
                        dp[i][k]=max(max(dp[i][k-c[j]]+g[j],dp[i][k]),dp[i-1][k-c[j]]+g[j]);
                    }
                }
                if(s==1)
                {
                    for(int j=0;j<=t;j++)
                    {
                        dp[i][j]=dp[i-1][j];
                    }
                    for(int j=1;j<=m;j++)
                    for(int k=t;k>=c[j];k--)
                    {
                        dp[i][k]=max(dp[i-1][k-c[j]]+g[j],dp[i][k]);
                    }
                }
                if(s==2)
                {
                    for(int j=0;j<=t;j++)
                    {
                        dp[i][j]=dp[i-1][j];
                    }
                    for(int j=1;j<=m;j++)
                    for(int k=t;k>=c[j];k--)
                    {
                        dp[i][k]=max(dp[i][k-c[j]]+g[j],dp[i][k]);
                    }
                }
            }
            dp[n][t]=max(dp[n][t],-1);
            printf("%d\n",dp[n][t]);
        }
        return 0;
    }