Hdu1078 memory-based search

Hdu1078 memory-based search

Question:

For n * n maps, each grid has a value that can only go from a small grid to a large grid. k can go up, down, and up to k steps each time to ask you about the maximum number of grids you have taken. what is

Dp [I] [j] indicates the maximum value that can be obtained starting from (I, j) and then deeply searches for the solution.

#include
 
  #include
  
   #include
   
    using namespace std
    ;int n
    ,k
    ,dp
    [110
    ][110
    ],cost
    [110
    ][110
    ];int dir
    [4
    ][2
    ]={0
    ,1
    ,0
    ,-1
    ,1
    ,0
    ,-1
    ,0
    };int mark
    [110
    ][110
    ];int max
    (int a
    ,int b
    ){    return a
    >b
    ?a
    :b
    ;}int dfs
    (int x
    ,int y
    ){    int i
    ,j
    ,sum
    =0
    ;    for(i
    =1
    ;i
    <=k
    ;i
    ++)    {        for(j
    =0
    ;j
    <4
    ;j
    ++)        {            int xx
    =x
    +dir
    [j
    ][0
    ]*i
    ;            int yy
    =y
    +dir
    [j
    ][1
    ]*i
    ;            if(xx
    <0
    ||xx
    >=n
    ||yy
    <0
    ||yy
    >=n
    ) continue;            if(cost
    [xx
    ][yy
    ]<=cost
    [x
    ][y
    ]) continue;            if(mark
    [xx
    ][yy
    ])            {                sum
    =max
    (sum
    ,dp
    [xx
    ][yy
    ]);            }            else             {                mark
    [xx
    ][yy
    ]=1
    ;                sum
    =max
    (sum
    ,dfs
    (xx
    ,yy
    ));            }        }    }    dp
    [x
    ][y
    ]=cost
    [x
    ][y
    ]+sum
    ;    return dp
    [x
    ][y
    ];    }int main(){    int i
    ,j
    ;    while(~scanf
    ("%d%d"
    ,&n
    ,&k
    ))    {        if(n
    <0
    ) break;        for(i
    =0
    ;i
    <n
    ;i
    ++)        for(j
    =0
    ;j
    <n
    ;j
    ++)        {            scanf
    ("%d"
    ,&cost
    [i
    ][j
    ]);        }        memset
    (dp
    ,0
    ,sizeof(dp
    ));        memset
    (mark
    ,0
    ,sizeof(mark
    ));        mark
    [0
    ][0
    ]=1
    ;        printf
    ("%d\n"
    ,dfs
    (0
    ,0
    ));    }    return 0
    ;}