Bugs Integrated, Inc.
| Time Limit: 15000MS | Memory Limit: 30000K | |
| Total Submissions: 9088 | Accepted: 3472 | |
| Case Time Limit: 5000MS | ||
Description
Bugs Integrated, Inc. is a major manufacturer of advanced memory chips. They are launching production of a new six terabyte Q-RAM chip. Each chip consists of six unit squares arranged in a form of a 2*3 rectangle. The way Q-RAM chips are made is such that one takes a rectangular plate of silicon divided into N*M unit squares. Then all squares are tested carefully and the bad ones are marked with a black marker. 
Finally, the plate of silicon is cut into memory chips. Each chip
consists of 2*3 (or 3*2) unit squares. Of course, no chip can contain
any bad (marked) squares. It might not be possible to cut the plate so
that every good unit square is a part of some memory chip. The
corporation wants to waste as little good squares as possible. Therefore
they would like to know how to cut the plate to make the maximum number
of chips possible.
Task
You are given the dimensions of several silicon plates and a list of
all bad unit squares for each plate. Your task is to write a program
that computes for each plate the maximum number of chips that can be cut
out of the plate.
Input
The
first line of the input file consists of a single integer D (1 <= D
<= 5), denoting the number of silicon plates. D blocks follow, each
describing one silicon plate. The first line of each block contains
three integers N (1 <= N <= 150), M (1 <= M <= 10), K (0
<= K <= MN) separated by single spaces. N is the length of the
plate, M is its height and K is the number of bad squares in the plate.
The following K lines contain a list of bad squares. Each line consists
of two integers x and y (1 <= x <= N, 1 <= y <= M)
?coordinates of one bad square (the upper left square has coordinates
[1, 1], the bottom right is [N,M]).
Output
For
each plate in the input file output a single line containing the
maximum number of memory chips that can be cut out of the plate.
Sample Input
2 6 6 5 1 4 4 6 2 2 3 6 6 4 6 5 4 3 3 6 1 6 2 6 4
Sample Output
3 4 题意简述： 一个生产芯片的工厂要生产一种规格为2X3的芯片，方法是先产生一块规格为nXm的矩形硅片，有nXm个正方形方块组成，但是硅片上存在一些损坏的方块，显示为黑色，它们不能用来制作芯片。现在给出硅片上每个损坏方块的位置，求用该硅片最多能切割出多少块芯片。 思路： 要知道能否存在第k-2行至第k行这3行中切割出芯片及切出多少芯片，需要知道这3行上每个方块的具体状态：是否损坏，是否已经被其他芯片使用，这两种情况都称为不可用。那么可以用一个三进制数来表示某一列第k-1至第k行的状态：0表示两个方块都可用，1表示仅有第k-1行的方块可用，2表示第k行的方块不可用。 用dp[k][t]表示第k-1行至第k行所有列的三进制特征数分别等于三进制数t的每一位的状态下最多能够切割出的芯片数。那么在状态转移时，凭借上一层的状态和本层的状态就能够知道这3行的全部情况。 该题中父状态和子状态都是多对多的关系，应该对每个可行的子状态（有些状态时不可用形成的），求切割父状态并更新它的最大值。
1 #include<cstdio>
2 int map[155][15];
3 int tri[12]={0,1,3,9,27,81,243,729,2187,6561,19683,59049};
4 int dp[2][59049];//用滚动数组记录每一行的状态
5 int pre[15],now[15];//上一行与本行的三进制状态
6 int n,m;
7 int max(int a,int b)
8 {
9 return (a>=b)?a:b;
10 }
11 int getten(int a[])//把三进制状态转化为十进制数
12 {
13 int sum=0;
14 for(int i=1;i<=m;++i)
15 sum+=tri[i]*a[i];
16 return sum;
17 }
18 void gettri(int k,int a[])//把十进制数转化为三进制数
19 {
20 for(int i=1;i<=m;i++)
21 {
22 a[i]=k%3;
23 k=k/3;
24 }
25 return;
26 }
27 void dfs(int i,int j,int last,int key)
28 {
29 int k;
30 dp[i%2][key]=max(dp[i%2][key],last);//不做改动，更新父状态
31 if(j>=m)
32 return;
33 if(!pre[j] && !pre[j+1] && !now[j] && !now[j+1])//竖着切割
34 {
35 now[j]=now[j+1]=2;
36 k=getten(now);
37 dfs(i,j+2,last+1,k);
38 now[j]=now[j+1]=0;
39 }
40 if(j<m-1 && !now[j] && !now[j+1] && !now[j+2])
41 {
42 now[j]=now[j+1]=now[j+2]=2;
43 k=getten(now);
44 dfs(i,j+3,last+1,k);
45 now[j]=now[j+1]=now[j+2]=0;
46 }
47 dfs(i,j+1,last,key);//不做改动，深搜下一列
48 return;
49 }
50 int main()
51 {
52 int nn,k,a,b,tmp,ans;
53 scanf("%d",&nn);
54 while(nn--)
55 {
56 scanf("%d %d %d",&n,&m,&k);
57 for(int i=0;i<tri[m+1];++i)
58 dp[1][i]=-1;//先初始化第一行的全部状态都是不可能形成的
59 for(int i=1;i<=m;++i)
60 for(int j=1;j<=n;++j)
61 map[j][i]=0;
62 for(int i=1;i<=k;++i)
63 {
64 scanf("%d %d",&a,&b);
65 map[a][b]=1;
66 }
67 for(int i=1;i<=m;++i)
68 pre[i]=map[1][i]+1;//计算第一行的状态第0行的方块全部视为不可用
69 tmp=getten(pre);
70 dp[1][tmp]=0;//设置第一行的原本状态时可以形成的，状态值为0
71 for(int i=2;i<=n;++i)//枚举子状态
72 {
73 for(int j=0;j<tri[m+1];++j)
74 dp[i%2][j]=-1;//先设置父状态值为-1，等待更新
75 for(int j=0;j<tri[m+1];++j)
76 {
77 if(dp[(i+1)%2][j]==-1)//跳过不可能的子状态
78 continue;
79 gettri(j,pre);
80 for(int t=1;t<=m;++t)
81 {
82 if(map[i][t])
83 now[t]=2;
84 else
85 now[t]=max(pre[t]-1,0);
86 }//根据上一行的状态得到本行的状态
87 tmp=getten(now);
88 dfs(i,1,dp[(i+1)%2][j],tmp);//深搜这一层所有可能的状态
89 }
90 }
91 ans=0;
92 for(int i=0;i<tri[m+1];++i)
93 ans=max(ans,dp[n%2][i]);//得到结果
94 printf("%d\n",ans);
95 }
96 return 0;
97 }