Hdu4374 monotonous queue optimized dp

Hdu4374 monotonous queue optimized dp

 

I won't talk much about the meaning of the question. Let's talk about the ideas directly;

There are two ways to reach each layer's point (T step on the left side or T step on the right side ).

 

Dp [I] [j] = max (dp [I] [j], dp [I-1] [k] + sum [I] [j]-sum [I] [k-1]); // From the top k to the right dp [I] [j] = max (dp [I] [j], dp [I-1] [k] + sum [I] [k]-sum [I] [j-1]); // Comes to the left from the upper k point

 

Because the sum of each layer is fixed, only the maximum value of the condition dp [I-1] [k]-sum [I] [k-1] is required. This uses the monotonic queue for optimization. now

 

 

#include
 
  #include
  
   #include
   
    using namespace std
    ;#define INF -0x3f3f3f3f
    int sum
    [20100
    ],num
    [110
    ][20100
    ],id
    [20100
    ],dp
    [110
    ][20100
    ],map
    [10010
    ];int max
    (int a
    ,int b
    ){    return a
    >b
    ?a
    :b
    ;}int main(){    int n
    ,m
    ,i
    ,j
    ,x
    ,T
    ;    while(~scanf
    ("%d%d%d%d"
    ,&n
    ,&m
    ,&x
    ,&T
    ))    {        for(i
    =1
    ;i
    <=n
    ;i
    ++)        for(j
    =1
    ;j
    <=m
    ;j
    ++)        {            scanf
    ("%d"
    ,&num
    [i
    ][j
    ]);            dp
    [i
    ][j
    ]=INF
    ;        }        dp
    [1
    ][x
    ]=num
    [1
    ][x
    ];        for(i
    =x
    -1
    ;i
    >=1
    &&x
    -i
    <=T
    ;i
    --)        dp
    [1
    ][i
    ]=dp
    [1
    ][i
    +1
    ]+num
    [1
    ][i
    ];        for(i
    =x
    +1
    ;i
    <=m
    &&i
    -x
    <=T
    ;i
    ++)        dp
    [1
    ][i
    ]=dp
    [1
    ][i
    -1
    ]+num
    [1
    ][i
    ];        int front
    ,top
    ;        sum
    [0
    ]=0
    ;        for(i
    =2
    ;i
    <=n
    ;i
    ++)        {            for(j
    =1
    ;j
    <=m
    ;j
    ++)            sum
    [j
    ]=sum
    [j
    -1
    ]+num
    [i
    ][j
    ];            front
    =0
    ,top
    =0
    ;            for(j
    =1
    ;j
    <=m
    ;j
    ++)            {                int tt
    =dp
    [i
    -1
    ][j
    ]-sum
    [j
    -1
    ];                while(front
    <top
    &&tt
    >map
    [top
    ])                top
    --;                id
    [++top
    ]=j
    ;                map
    [top
    ]=tt
    ;                while(j
    -T
    >id
    [front
    +1
    ]&&front
    <top
    )                front
    ++;                dp
    [i
    ][j
    ]=max
    (dp
    [i
    ][j
    ],map
    [front
    +1
    ]+sum
    [j
    ]);                }            front
    =0
    ,top
    =0
    ;            sum
    [m
    +1
    ]=0
    ;            for(j
    =m
    ;j
    >=1
    ;j
    --)            sum
    [j
    ]=sum
    [j
    +1
    ]+num
    [i
    ][j
    ];            for(j
    =m
    ;j
    >=1
    ;j
    --)            {                int tt
    =dp
    [i
    -1
    ][j
    ]-sum
    [j
    +1
    ];                while(front
    <top
    &&tt
    >map
    [top
    ])                top
    --;                id
    [++top
    ]=j
    ;                map
    [top
    ]=tt
    ;                while(front
    <top
    &&id
    [front
    +1
    ]>j
    +T
    )                front
    ++;                dp
    [i
    ][j
    ]=max
    (dp
    [i
    ][j
    ],map
    [front
    +1
    ]+sum
    [j
    ]);            }        }        int Max
    =INF
    ;        for(i
    =1
    ;i
    <=m
    ;i
    ++)        if(dp
    [n
    ][i
    ]>Max
    ) Max
    =dp
    [n
    ][i
    ];        printf
    ("%d\n"
    ,Max
    );    }    return 0
    ;}