HDU-3449-Consumer（有依赖背包）

骑猪看日落 2022-05-26 13:45 120阅读 0赞

# Consumer #

#### Problem Description ####

FJ is going to do some shopping, and before that, he needs some boxes to carry the different kinds of stuff he is going to buy. Each box is assigned to carry some specific kinds of stuff (that is to say, if he is going to buy one of these stuff, he has to buy the box beforehand). Each kind of stuff has its own value. Now FJ only has an amount of W dollars for shopping, he intends to get the highest value with the money.

#### Input ####

The first line will contain two integers, n (the number of boxes 1 <= n <= 50), w (the amount of money FJ has, 1 <= w <= 100000) Then n lines follow. Each line contains the following number pi (the price of the ith box 1<=pi<=1000), mi (1<=mi<=10 the number goods ith box can carry), and mi pairs of numbers, the price cj (1<=cj<=100), the value vj(1<=vj<=1000000)

#### Output ####

For each test case, output the maximum value FJ can get

#### Sample Input ####

3 800  
300 2 30 50 25 80  
600 1 50 130  
400 3 40 70 30 40 35 60

#### Sample Output ####

210

**题目意思：**有N个箱子，每个箱子需要花费Ai的代价

箱子里面有K个物品，K个物品你可以选择买任意个，但每个物品只能买一个，每个物品有相应的花费和价值

你又M的大洋，最后问最多能得到多少价值物品。

**解题思路：**先对每组选择里的所有物品进行0-1背包处理，但背包容量为（总容量\-盒子容量）；然后跟上一组的状态比较来决定这一组选择是选还是不选，取其中的较大值。

#include<bits/stdc++.h>
    using namespace std;
    int dp[100005];
    int tmp[100005];
    int main()
    {
        int n,w,p,m,c,v;
        while(~scanf("%d%d",&n,&w))
        {
            memset(dp,0,sizeof(dp));
            for(int i=0;i<n;i++)
            {
                memcpy(tmp,dp,sizeof(dp));///继承前面的
                scanf("%d%d",&p,&m);
                for(int j=0;j<m;j++)
                {
                    scanf("%d%d",&c,&v);
                    for(int k=w-p;k>=c;k--)
                    {
                        tmp[k]=max(tmp[k],tmp[k-c]+v);
                    }
                }
                for(int k=p;k<=w;k++)
                {
                    dp[k]=max(dp[k],tmp[k-p]);
                }
            }
            printf("%d\n",dp[w]);
        }
        return 0;
    }