（状压dp）HDU 4778 Gems Fight!

Gems Fight!

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 327680/327680 K (Java/Others)
Total Submission(s): 2395    Accepted Submission(s): 1087

Problem Description

　　Alice and Bob are playing "Gems Fight!":
　　There are Gems of G different colors , packed in B bags. Each bag has several Gems. G different colors are numbered from color 1 to color G.
　　Alice and Bob take turns to pick one bag and collect all the Gems inside. A bag cannot be picked twice. The Gems collected are stored in a shared cooker.
　　After a player ,we name it as X, put Gems into the cooker, if there are S Gems which are the same color in the cooker, they will be melted into one Magic Stone. This reaction will go on and more than one Magic Stone may be produced, until no S Gems of the same color remained in that cooker. Then X owns those new Magic Stones. When X gets one or more new Magic Stones, he/she will also get a bonus turn. If X gets Magic Stone in a bonus turn, he will get another bonus turn. In short,a player may get multiple bonus turns continuously.
　　There will be B turns in total. The goal of "Gems Fight!" is to get as more Magic Stones than the opponent as possible.
　　Now Alice gets the first turn, and she wants to know, if  both of them act the optimal way, what will be the difference between the number of her Magic Stones and the number of Bob's Magic Stones at the end of the game.

 

 

Input

　　There are several cases(<=20).
　　In each case, there are three integers at the first line: G, B, and S. Their meanings are mentioned above.
　　Then B lines follow. Each line describes a bag in the following format:
　　
　　n c ₁ c ₂ ... c _n
　　
　　It means that there are n Gems in the bag and their colors are color c ₁,color c ₂...and color c _n respectively.
　　 0<=B<=21, 0<=G<=8, 0<n<=10, S < 20.
　　There may be extra blank lines between cases. You can get more information from the sample input.
　　The input ends with G = 0, B = 0 and S = 0.

 

 

Output

　　One line for each case: the amount of Alice's Magic stones minus the amount of Bob's Magic Stones.

 

 

Sample Input

3 4 3 2 2 3 2 1 3 2 1 2 3 2 3 1 3 2 2 3 2 3 1 3 1 2 3 0 0 0

 

 

Sample Output

3 -3

Hint

　　For the first case, in turn 2, bob has to choose at least one bag, so that Alice will make a Magic Stone at the end of turn 3, thus get turn 4 and get all the three Magic Stones.

 

 

Source

2013 Asia Hangzhou Regional Contest

 

用dp[st]表示状态st下，用余下的未选的包裹能对当前先手做出的最大贡献。（即使当下的先手最大化其最终个数与对手的个数的差值）。末状态时显然贡献为0（双方均无包裹可再取）。而最初的的dp值显然需要由其产生的各种状况综合比较得到，故我们考虑逆向dp。每次先求出当前状态各种g余下的个数，对于下一个状况，统计加上后g的个数，设所有新的g共可以生成cnt个魔法石，如果cnt>0，则下一个状况的先手存在就是当下状况先手的可能，故用cnt+dp[st_new]更新，不然必改变了先手，用cnt-dp[st_new]更新。

 1 #include <iostream>
 2 #include <string>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <cstdio>
 6 #include <cmath>
 7 #include <queue>
 8 #include <set>
 9 #include <map>
10 #include <list>
11 #include <vector>
12 #include <stack>
13 #define mp make_pair
14 //#define P make_pair
15 #define MIN(a,b) (a>b?b:a)
16 //#define MAX(a,b) (a>b?a:b)
17 typedef long long ll;
18 typedef unsigned long long ull;
19 const int MAX=40;
20 const int INF=1e9+5;
21 using namespace std;
22 //const int MOD=1e9+7;
23 typedef pair<ll,int> pii;
24 const double eps=0.00000001;
25 int dp[1<<21];
26 int num[30][10];
27 int yu[10];//每种余下的数目
28 int trial[10];
29 int b,g,n,s,tem,cnt;
30 int main()
31 {
32     while(scanf("%d%d%d",&g,&b,&s)&&!(g==0&&b==0&&s==0))
33     {
34         memset(dp,0,sizeof(dp));
35         memset(num,0,sizeof(num));
36         for(int i=0;i<b;i++)
37         {
38             scanf("%d",&n);
39             for(int j=1;j<=n;j++)
40             {
41                 scanf("%d",&tem);
42                 ++num[i][tem];
43             }
44         }
45         int total=(1<<b)-1;
46         for(int i=1;i<=total;i++)
47         {
48             dp[i]=-INF;
49             memset(yu,0,sizeof(yu));
50             for(int j=0;j<b;j++)
51             {
52                 if(!(i&(1<<j)))
53                 {
54                     for(int q=1;q<=g;q++)
55                         yu[q]+=num[j][q];
56                 }
57             }
58             for(int q=1;q<=g;q++)
59                 yu[q]%=s;
60             for(int j=0;j<b;j++)
61             {
62                 if(i&(1<<j))
63                 {
64                     cnt=0;
65                     memcpy(trial,yu,sizeof(yu));
66                     for(int q=1;q<=g;q++)
67                     {
68                         trial[q]+=num[j][q];
69                         while(trial[q]>=s)
70                         {
71                             ++cnt;
72                             trial[q]-=s;
73                         }
74                     }
75                     if(cnt)
76                         dp[i]=max(dp[i],dp[i^(1<<j)]+cnt);
77                     else
78                         dp[i]=max(dp[i],-dp[i^(1<<j)]+cnt);
79                 }
80             }
81         }
82         printf("%d\n",dp[total]);
83     }
84 }

 

转载于:https://www.cnblogs.com/quintessence/p/7242342.html