uva 103 - Stacking Boxes

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=114&page=show_problem&problem=39

 

题目读了好久总算看懂了，具有n维属性的箱子a和b，如果a fits b 仅当存在a的属性排列，使与b对应

的属性满足q(a)<q(b)，算法中定义二维数组，标记a是否fits b，然后深搜，当然得先把每个箱子的属性排序。

 

  
    

   
     
   
      1 
   
     #include 
   
     <
   
     stdio.h
   
     >
   
     

   
      2 
   
     #include 
   
     <
   
     string
   
     .h
   
     >
   
     

   
      3 
   
     #include 
   
     <
   
     stdlib.h
   
     >
   
     

   
      4 
   
     

   
      5 
   
      
   
     struct
   
      Box {

   
      6 
   
      
   
     int
   
      mt[
   
     15
   
     ];

   
      7 
   
     }bx[
   
     40
   
     ];

   
      8 
   
      
   
     bool
   
      flg[
   
     40
   
     ][
   
     40
   
     ], f0[
   
     40
   
     ];

   
      9 
   
      
   
     int
   
      rst[
   
     40
   
     ], tp[
   
     40
   
     ];

   
     10 
   
     
   
     int
   
      nb, mt;

   
     11 
   
     
   
     int
   
      max;

   
     12 
   
     

   
     13 
   
     
   
     int
   
      comp(
   
     const
   
      
   
     void
   
      
   
     *
   
     a, 
   
     const
   
      
   
     void
   
      
   
     *
   
     b)

   
     14 
   
     {

   
     15 
   
      
   
     return
   
      
   
     *
   
     ((
   
     int
   
      
   
     *
   
     )a) 
   
     -
   
      
   
     *
   
     ((
   
     int
   
      
   
     *
   
     )b);

   
     16 
   
     }

   
     17 
   
     

   
     18 
   
     
   
     bool
   
      fits(
   
     int
   
      i, 
   
     int
   
      j)

   
     19 
   
     {

   
     20 
   
      
   
     for
   
      (
   
     int
   
      k
   
     =
   
     0
   
     ; k
   
     <
   
     mt; k
   
     ++
   
     )

   
     21 
   
      
   
     if
   
      (bx[i].mt[k] 
   
     >=
   
      bx[j].mt[k]) 
   
     return
   
      
   
     false
   
     ;

   
     22 
   
      
   
     return
   
      
   
     true
   
     ;

   
     23 
   
     }

   
     24 
   
     

   
     25 
   
     
   
     void
   
      DFS(
   
     int
   
      i, 
   
     int
   
      t)

   
     26 
   
     {

   
     27 
   
      
   
     int
   
      j;

   
     28 
   
      
   
     for
   
      (j
   
     =
   
     1
   
     ; j
   
     <=
   
     nb; j
   
     ++
   
     ) {

   
     29 
   
      
   
     if
   
      (flg[i][j]) {

   
     30 
   
      tp[t
   
     +
   
     1
   
     ] 
   
     =
   
      j;

   
     31 
   
      DFS(j, t
   
     +
   
     1
   
     );

   
     32 
   
      }

   
     33 
   
      }

   
     34 
   
      
   
     if
   
      (max 
   
     <
   
      t) {

   
     35 
   
      
   
     for
   
      (j
   
     =
   
     1
   
     ; j
   
     <=
   
     t; j
   
     ++
   
     )

   
     36 
   
      rst[j] 
   
     =
   
      tp[j];

   
     37 
   
      max 
   
     =
   
      t;

   
     38 
   
      }

   
     39 
   
     }

   
     40 
   
     

   
     41 
   
     
   
     int
   
      main()

   
     42 
   
     {

   
     43 
   
      
   
     int
   
      i, j;

   
     44 
   
      
   
     while
   
      (scanf(
   
     "
   
     %d%d
   
     "
   
     , 
   
     &
   
     nb, 
   
     &
   
     mt) 
   
     !=
   
      EOF) {

   
     45 
   
      
   
     for
   
      (i
   
     =
   
     1
   
     ; i
   
     <=
   
     nb; i
   
     ++
   
     ) {

   
     46 
   
      
   
     for
   
      (j
   
     =
   
     0
   
     ; j
   
     <
   
     mt; j
   
     ++
   
     )

   
     47 
   
      scanf(
   
     "
   
     %d
   
     "
   
     , 
   
     &
   
     bx[i].mt[j]);

   
     48 
   
      qsort(bx[i].mt, mt, 
   
     sizeof
   
     (bx[i].mt[
   
     0
   
     ]), comp);

   
     49 
   
      }

   
     50 
   
      memset(flg, 
   
     0
   
     , 
   
     sizeof
   
     (flg));

   
     51 
   
      
   
     for
   
      (i
   
     =
   
     1
   
     ; i
   
     <=
   
     nb; i
   
     ++
   
     ) {

   
     52 
   
      
   
     for
   
      (j
   
     =
   
     1
   
     ; j
   
     <=
   
     nb; j
   
     ++
   
     ) {

   
     53 
   
      
   
     if
   
      (i 
   
     ==
   
      j) 
   
     continue
   
     ;

   
     54 
   
      
   
     if
   
      (fits(i, j)) {

   
     55 
   
      flg[i][j] 
   
     =
   
      
   
     1
   
     ;

   
     56 
   
      }

   
     57 
   
      }

   
     58 
   
      }

   
     59 
   
      max 
   
     =
   
      
   
     0
   
     ;

   
     60 
   
      

   
     61 
   
      
   
     for
   
      (i
   
     =
   
     1
   
     ; i
   
     <=
   
     nb; i
   
     ++
   
     ) {

   
     62 
   
      tp[
   
     1
   
     ] 
   
     =
   
      i;

   
     63 
   
      memset(f0, 
   
     0
   
     , 
   
     sizeof
   
     (f0));

   
     64 
   
      f0[i] 
   
     =
   
      
   
     1
   
     ;

   
     65 
   
      DFS(i, 
   
     1
   
     );

   
     66 
   
      }

   
     67 
   
      
   
     //
   
     if a fits b , b fits c then a fits c

   
     68 
   
      
   
     //
   
     因此rst[]保存的就是期望的顺序
   
     

   
     69 
   
     
   
      printf(
   
     "
   
     %d\n
   
     "
   
     , max);

   
     70 
   
     

   
     71 
   
      
   
     for
   
     (i
   
     =
   
     1
   
     ; i
   
     <=
   
     max; i
   
     ++
   
     ) {

   
     72 
   
      printf(
   
     "
   
     %d
   
     "
   
     , rst[i]);

   
     73 
   
      
   
     if
   
      (i 
   
     !=
   
      max) printf(
   
     "
   
      
   
     "
   
     );

   
     74 
   
      }

   
     75 
   
      printf(
   
     "
   
     \n
   
     "
   
     );

   
     76 
   
      }

   
     77 
   
      
   
     return
   
      
   
     0
   
     ;

   
     78 
   
     }

   
     79