Hamburger Magi(hdu 3182)

Hamburger Magi

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 437    Accepted Submission(s): 144

Problem Description

In the mysterious forest, there is a group of Magi. Most of them like to eat human beings, so they are called “The Ogre Magi”, but there is an special one whose favorite food is hamburger, having been jeered by the others as “The Hamburger Magi”.
Let’s give The Hamburger Magi a nickname “HamMagi”, HamMagi don’t only love to eat but also to make hamburgers, he makes N hamburgers, and he gives these each hamburger a value as Vi, and each will cost him Ei energy, (He can use in total M energy each day). In addition, some hamburgers can’t be made directly, for example, HamMagi can make a “Big Mac” only if “New Orleams roasted burger combo” and “Mexican twister combo” are all already made. Of course, he will only make each kind of hamburger once within a single day. Now he wants to know the maximal total value he can get after the whole day’s hard work, but he is too tired so this is your task now!

 

Input

The first line consists of an integer C(C<=50), indicating the number of test cases.
The first line of each case consists of two integers N,E(1<=N<=15,0<=E<=100) , indicating there are N kinds of hamburgers can be made and the initial energy he has.
The second line of each case contains N integers V1,V2…VN, (Vi<=1000)indicating the value of each kind of hamburger.
The third line of each case contains N integers E1,E2…EN, (Ei<=100)indicating the energy each kind of hamburger cost.
Then N lines follow, each line starts with an integer Qi, then Qi integers follow, indicating the hamburgers that making ith hamburger needs.

 

Output

For each line, output an integer indicating the maximum total value HamMagi can get.

 

Sample Input

1

4 90

243 464 307 298

79 58 0 72

3 2 3 4

2 1 4

1 1

0

Sample Output

298

题意：就是给你n块面包，给出他的价值以及所需消耗体力。

给出人物的体力，然后每个面包做需要一定的条件，就是在这个面包做前其他几个面包要做好。

问最后得到的价值最大。

思路：状压dp;

总共pow(2,n)的状态。

dp[i]表示在状态i所得到的价值的最大。

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<string.h>
 5 #include<stdlib.h>
 6 #include<queue>
 7 #include<stack>
 8 using namespace std;
 9 typedef struct qq
10 {
11     int x;//记入当前状态最大价值
12     int y;//记录剩余的体力
13 } mm;
14 typedef struct pp
15 {
16     int x;
17     int y;
18     int cnt;
19     int a[20];
20 } ss;//记入面包的价值消耗以及要求
21 ss kk[22];
22 mm dp[1<<16];
23 int main(void)
24 {
25     int n,i,j,k,p,q;
26     scanf("%d",&k);
27     while(k--)
28     {
29         scanf("%d %d",&p,&q);
30         for(i=0; i<p; i++)
31         {
32             scanf("%d",&kk[i].x);
33         }
34         for(i=0; i<p; i++)
35         {
36             scanf("%d",&kk[i].y);
37         }
38         for(i=0; i<p; i++)
39         {
40             scanf("%d",&kk[i].cnt);
41             for(j=0; j<kk[i].cnt; j++)
42             {
43                 scanf("%d",&kk[i].a[j]);
44                 kk[i].a[j]-=1;
45             }
46         }
47         for(i=0; i<(1<<16); i++)
48         {
49             dp[i].x=0;
50             dp[i].y=-100;
51         }
52         int maxx=0;
53         dp[0].y=q;//初始化
54         for(i=1; i<(1<<p); i++)
55         {
56             for(j=0; j<p; j++)
57             {
58                 if(i&(1<<j))//判断是否在当前的状态下
59                 {
60                     int c=i^(1<<j);//找这个状态的前一个状态
61                     int s=0;
62                     for(s=0; s<kk[j].cnt; s++)//判断要求是否成立也就是前一个状态下是否有做这个面包的要求
63                     {
64                         if((c&(1<<(kk[j].a[s])))==0)
65                         {
66                             break;
67                         }
68                     }
69                     if(s==kk[j].cnt)
70                     {
71                         if(dp[c].y>=kk[j].y)
72                         {
73                             int cc=dp[c].x+kk[j].x;
74                             if(cc>dp[i].x)
75                             {
76                                 dp[i].x=cc;
77                                 dp[i].y=dp[c].y-kk[j].y;
78                             }
79                             if(dp[i].x>maxx)
80                             {
81                                 maxx=dp[i].x;//更新最大值
82                             }
83                         }
84                     }
85                 }
86
87             }
88
89         }
90         printf("%d\n",maxx);
91     }
92     return 0;
93 }

状压DP