UVa 437 - The Tower of Babylon (White Book)

题目地址：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=6&problem=378&mosmsg=Submission+received+with+ID+17127418

思路：用2A dimension represents the state transition,这个思路挺巧妙的

AC代码：

登录后复制

      
      

       
       #include <iostream>
       
       

       
       #include <cstdio>
       
       

       
       #include <cstdlib>
       
       

       
       #include <algorithm>
       
       

       
       #include <queue>
       
       

       
       #include <stack>
       
       

       
       #include <map>
       
       

       
       #include <cstring>
       
       

       
       #include <climits>
       
       

       
       #include <cmath>
       
       

       
       #include <cctype>
       
       

       
       const 
       
       int 
       
       inf 
       
       = 
       
       0x3f3f3f3f;
       
       //1061109567
       
       

       
       typedef 
       
       long 
       
       long 
       
       LL;
       
       

       
       #define lson l,m,rt<<1
       
       

       
       #define rson m+1,r,rt<<1|1
       
       

       
       using 
       
       namespace 
       
       std;
       
       

       
       int 
       
       value[
       
       35][
       
       3];
       
       //The first dimension represents the ordinal number of the cube,The second dimension represents the length of the cube,宽,高
       
       

       
       int 
       
       dp1[
       
       35][
       
       3];
       
       

       
       int 
       
       n;
       
       

       
       void 
       
       get_arr(
       
       int
       
       * 
       
       cf,
       
       int 
       
       a,
       
       int 
       
       b)
       
       
{
       
       
    
       
       int 
       
       k 
       
       = 
       
       0;
       
       
    
       
       for(
       
       int 
       
       i
       
       =
       
       0; 
       
       i
       
       <
       
       3; 
       
       i
       
       ++)
       
       
    {
       
       
        
       
       if(
       
       i 
       
       != 
       
       b)
       
       
            
       
       cf[
       
       k
       
       ++] 
       
       = 
       
       value[
       
       a][
       
       i];
       
       
    }
       
       
}
       
       

       
       int 
       
       dp(
       
       int 
       
       x,
       
       int 
       
       y)
       
       //用iIndicates the serial number of the cube,jIndicates the height of the cube,to represent a state
       
       
{
       
       
    
       
       int 
       
       b[
       
       2],
       
       c[
       
       2];
       
       //这2It is wrong to open an array as a global
       
       
    
       
       if(
       
       dp1[
       
       x][
       
       y] 
       
       > 
       
       0)
       
       
        
       
       return 
       
       dp1[
       
       x][
       
       y];
       
       
    
       
       int 
       
       ans 
       
       = 
       
       0;
       
       
    
       
       get_arr(
       
       b,
       
       x,
       
       y);
       
       
    
       
       for(
       
       int 
       
       i
       
       =
       
       0; 
       
       i
       
       <
       
       n; 
       
       i
       
       ++)
       
       
    {
       
       
        
       
       for(
       
       int 
       
       j
       
       =
       
       0; 
       
       j
       
       <
       
       3; 
       
       j
       
       ++)
       
       
        {
       
       
            
       
       get_arr(
       
       c,
       
       i,
       
       j);
       
       
            
       
       if(
       
       b[
       
       0] 
       
       > 
       
       c[
       
       0] 
       
       && 
       
       b[
       
       1] 
       
       > 
       
       c[
       
       1])
       
       
                
       
       ans 
       
       = 
       
       max(
       
       ans,
       
       dp(
       
       i,
       
       j));
       
       
        }
       
       
    }
       
       
    
       
       ans 
       
       += 
       
       value[
       
       x][
       
       y];
       
       
    
       
       dp1[
       
       x][
       
       y] 
       
       = 
       
       ans;
       
       
    
       
       return 
       
       ans;
       
       
}
       
       

       
       int 
       
       main()
       
       
{
       
       
    
       
       int 
       
       cas 
       
       = 
       
       1;
       
       
    
       
       while(
       
       scanf(
       
       "%d",
       
       &
       
       n) 
       
       && 
       
       n)
       
       
    {
       
       
        
       
       memset(
       
       dp1,
       
       0,
       
       sizeof(
       
       dp1));
       
       
        
       
       for(
       
       int 
       
       i
       
       =
       
       0; 
       
       i
       
       <
       
       n; 
       
       i
       
       ++)
       
       
        {
       
       
            
       
       for(
       
       int 
       
       j
       
       =
       
       0; 
       
       j
       
       <
       
       3; 
       
       j
       
       ++)
       
       
                
       
       scanf(
       
       "%d",
       
       &
       
       value[
       
       i][
       
       j]);
       
       
            
       
       sort(
       
       value[
       
       i],
       
       value[
       
       i]
       
       +
       
       3);
       
       
        }
       
       
        
       
       int 
       
       ans 
       
       = 
       
       0;
       
       
        
       
       for(
       
       int 
       
       i
       
       =
       
       0; 
       
       i
       
       <
       
       n; 
       
       i
       
       ++)
       
       //Enumerate the bottom case of each cube
       
       
        {
       
       
            
       
       for(
       
       int 
       
       j
       
       =
       
       0; 
       
       j
       
       <
       
       3; 
       
       j
       
       ++)
       
       
                
       
       ans 
       
       = 
       
       max(
       
       ans,
       
       dp(
       
       i,
       
       j));
       
       
        }
       
       
        
       
       printf(
       
       "Case %d: maximum height = %d\n",
       
       cas
       
       ++,
       
       ans);
       
       
    }
       
       
    
       
       return 
       
       0;
       
       
}
      
      
      
      

       
       
1.
       
       
2.
       
       
3.
       
       
4.
       
       
5.
       
       
6.
       
       
7.
       
       
8.
       
       
9.
       
       
10.
       
       
11.
       
       
12.
       
       
13.
       
       
14.
       
       
15.
       
       
16.
       
       
17.
       
       
18.
       
       
19.
       
       
20.
       
       
21.
       
       
22.
       
       
23.
       
       
24.
       
       
25.
       
       
26.
       
       
27.
       
       
28.
       
       
29.
       
       
30.
       
       
31.
       
       
32.
       
       
33.
       
       
34.
       
       
35.
       
       
36.
       
       
37.
       
       
38.
       
       
39.
       
       
40.
       
       
41.
       
       
42.
       
       
43.
       
       
44.
       
       
45.
       
       
46.
       
       
47.
       
       
48.
       
       
49.
       
       
50.
       
       
51.
       
       
52.
       
       
53.
       
       
54.
       
       
55.
       
       
56.
       
       
57.
       
       
58.
       
       
59.
       
       
60.
       
       
61.
       
       
62.
       
       
63.
       
       
64.
       
       
65.
       
       
66.
       
       
67.
       
       
68.
       
       
69.
       
       
70.