初次dp

http://acm.pku.edu.cn/JudgeOnline/problem?id=1163

这是我第一次用动态规划来做题。很不熟练啊。

从下往上依次考察（i，j）下面的较大数，并储存起来。

  
    

   
     
   
     #include
   
     <
   
     stdio.h
   
     >
   
     
#include
   
     <
   
     string
   
     .h
   
     >
   
     

   
     int
   
      n,input[
   
     100
   
     ][
   
     100
   
     ];

   
     long
   
      storage[
   
     100
   
     ][
   
     100
   
     ];

   
     int
   
      d(
   
     int
   
      i,
   
     int
   
      j)
{
 
   
     if
   
     (storage[i][j]
   
     >=
   
     0
   
     )
 
   
     return
   
      storage[i][j];
 
   
     return
   
      storage[i][j]
   
     =
   
     input[i][j]
   
     +
   
     (i
   
     ==
   
     (n
   
     -
   
     1
   
     )
   
     ?
   
     0
   
     :(storage[i
   
     +
   
     1
   
     ][j]
   
     >
   
     storage[i
   
     +
   
     1
   
     ][j
   
     +
   
     1
   
     ]
   
     ?
   
     storage[i
   
     +
   
     1
   
     ][j]:storage[i
   
     +
   
     1
   
     ][j
   
     +
   
     1
   
     ]));
}

   
     int
   
      main()
{
 
   
     int
   
      i,j;
 
   
     while
   
     (scanf(
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     n)
   
     !=
   
     EOF)
 {
 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )
 
   
     for
   
     (j
   
     =
   
     0
   
     ;j
   
     <=
   
     i;j
   
     ++
   
     )
 scanf(
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     input[i][j]);
 memset(storage,
   
     -
   
     1
   
     ,
   
     sizeof
   
     (storage));
 
   
     for
   
     (i
   
     =
   
     n
   
     -
   
     1
   
     ;i
   
     >=
   
     0
   
     ;i
   
     --
   
     )
 
   
     for
   
     (j
   
     =
   
     0
   
     ;j
   
     <=
   
     i;j
   
     ++
   
     )
 d(i,j);
 printf(
   
     "
   
     %ld\n
   
     "
   
     ,storage[
   
     0
   
     ][
   
     0
   
     ]);
 }
 
   
     return
   
      
   
     0
   
     ;
}
  
    

再次写这个代码：

  
    

   
     
   
     #include
   
     <
   
     stdio.h
   
     >
   
     


   
     int
   
      dp[
   
     350
   
     ][
   
     350
   
     ];

   
     int
   
      num[
   
     350
   
     ][
   
     350
   
     ];

   
     int
   
      n;


   
     int
   
      DP(
   
     int
   
      i, 
   
     int
   
      j)
{
 
   
     if
   
     (dp[i][j] 
   
     >=
   
      
   
     0
   
     )
 
   
     return
   
      dp[i][j];
 
   
     return
   
      dp[i][j] 
   
     =
   
      num[i][j] 
   
     +
   
      ((i
   
     +
   
     1
   
      
   
     <
   
      n)
   
     ?
   
     (DP(i
   
     +
   
     1
   
     , j) 
   
     >
   
      DP(i
   
     +
   
     1
   
     , j
   
     +
   
     1
   
     ) 
   
     ?
   
      DP(i
   
     +
   
     1
   
     , j) : DP(i
   
     +
   
     1
   
     , j
   
     +
   
     1
   
     )):
   
     0
   
     );
 
   
     //
   
     第一个括号少了，优先级问题
   
     

   
     }


   
     int
   
      main()
{
 
   
     int
   
      i, j;
 
   
     while
   
     (scanf(
   
     "
   
     %d
   
     "
   
     , 
   
     &
   
     n) 
   
     !=
   
      EOF)
 {
 
   
     for
   
     (i
   
     =
   
     0
   
     ; i
   
     <
   
     n; i
   
     ++
   
     )
 
   
     for
   
     (j
   
     =
   
     0
   
     ; j
   
     <=
   
     i; j
   
     ++
   
     )
 {
 scanf(
   
     "
   
     %d
   
     "
   
     , 
   
     &
   
     num[i][j]);
 dp[i][j] 
   
     =
   
      
   
     -
   
     1
   
     ; 
   
     //
   
     用了0， 在DP里判断用 if(dp[i][j] > 0) 超时
   
     

   
      }

 
   
     for
   
     (i
   
     =
   
     n
   
     -
   
     1
   
     ; i
   
     >=
   
     0
   
     ; i
   
     --
   
     )
 
   
     for
   
     (j
   
     =
   
     0
   
     ; j
   
     <=
   
     i; j
   
     ++
   
     )
 DP(i, j);

 printf(
   
     "
   
     %d\n
   
     "
   
     , dp[
   
     0
   
     ][
   
     0
   
     ]);
 }
 
   
     return
   
      
   
     0
   
     ;
}