Hdu 3450 tree array optimized dp

Source: Internet
Author: User

Hdu 3450 tree array optimized dp

 

I won't talk about the meaning of the question; hdu2227 is almost just that this question is not incremental, but the adjacent difference is not greater than d. In fact, they are the same and there are also differences;

This question does not contain a range and the size relationship between trees has an impact on the results. Therefore, the original id is used in the tree array. Due to the large number of values, the discretization feeling here is similar to the ing, discretization is used to find IDs;

When the id of num [I] is x, dp [I] indicates that the number of solutions is obvious dp [I] = sun (dp [j], j

 

#include
 
  #include#include
  
   #include
   
    using namespace std
    ;#define mod 9901
    int num1
    [100010
    ],num2
    [100010
    ],mark
    [100010
    ],tt
    ,n
    ;int coun
    [100010
    ];int update
    (int a
    ,int b
    ){    for(int i
    =a
    ;i
    <=n
    ;i
    +=(i
    &-i
    ))    {        coun
    [i
    ]+=b
    ;        coun
    [i
    ]%=mod
    ;    }    return 0
    ;}int find
    (int a
    ){    int s
    =0
    ;    for(int i
    =a
    ;i
    >=1
    ;i
    -=(i
    &-i
    ))    {        s
    +=coun
    [i
    ];        s
    %=mod
    ;    }    return s
    ;}int searchr
    (int a
    ){    int left
    =1
    ,right
    =tt
    ,now
    ;    while(left
    <=right
    )    {        int mid
    =(left
    +right
    )/2
    ;        if(mark
    [mid
    ]<=a
    )            {               now
    =mid
    ;            left
    =mid
    +1
    ;        }        else         {            right
    =mid
    -1
    ;        }    }        return now
    ;}int searchl
    (int a
    ){    int left
    =1
    ,right
    =tt
    ;    while(left
    <right
    )    {         int mid
    =(left
    +right
    )/2
    ;        if(mark
    [mid
    ]<a
    ) left
    =mid
    +1
    ;        else if(mark
    [mid
    ]==a
    ) return mid
    ;        else right
    =mid
    ;    }    return left
    ;}int main(){    int i
    ,j
    ,d
    ;    while(~scanf
    ("%d%d"
    ,&n
    ,&d
    ))    {        for(i
    =1
    ;i
    <=n
    ;i
    ++)        {            scanf
    ("%d"
    ,&num1
    [i
    ]);            num2
    [i
    ]=num1
    [i
    ];        }        sort
    (num2
    +1
    ,num2
    +1
    +n
    );        tt
    =0
    ;                 num2
    [0
    ]=-1
    ;        for(i
    =1
    ;i
    <=n
    ;i
    ++)        {            if(num2
    [i
    ]!=num2
    [i
    -1
    ])            {                tt
    ++;                mark
    [tt
    ]=num2
    [i
    ];            }        }        memset
    (coun
    ,0
    ,sizeof(coun
    ));        int sum
    =0
    ;        int x
    =searchl
    (num1
    [1
    ]);        for(i
    =1
    ;i
    <=n
    ;i
    ++)        {            int left
    =searchl
    (num1
    [i
    ]-d
    )-1
    ;            int right
    =searchr
    (num1
    [i
    ]+d
    );            int now
    =searchl
    (num1
    [i
    ]);            int s
    =find
    (right
    )-find
    (left
    );            s
    =(s
    %mod
    +mod
    )%mod
    ;            sum
    +=s
    ;            sum
    %=mod
    ;            update
    (now
    ,s
    +1
    );        }        printf
    ("%d\n"
    ,sum
    );    }    return 0
    ;}
   
  
 

 

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