CF 505C (Mr. Kitayuta, the Treasure Hunter-Dp considers the available range)

Source: Internet
Author: User

CF 505C (Mr. Kitayuta, the Treasure Hunter-Dp considers the available range)

C. Mr. Kitayuta, the Treasure Huntertime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

The Shuseki Islands are an archipelago of 30001 small islands in the Yutampo Sea. the islands are evenly spaced along a line and numbered from 0 to 30000 from the west to the east. these islands are known to contain your treasures. there areNGems in the Shuseki Islands in total, andI-Th gem is located on islandPI.

Mr. Kitayuta has just arrived at island 0. With his great jumping ability, he will repeatedly perform jumps between islands to the east according to the following process:

  • First, he will jump from island 0 to islandD.
  • After that, he will continue jumping according to the following rule. LetLBe the length of the previous jump, that is, if his previous jump was from islandPrevTo islandCur, LetL? =?Cur? -?Prev. He will perform a jump of lengthL? -? 1,LOrL? +? 1 to the east. That is, he will jump to island (Cur? +?L? -? 1 ),(Cur? +?L) Or (Cur? +?L? +? 1) (if they exist). The length of a jump must be positive, that is, he cannot perform a jump of length 0 whenL? =? 1. If there is no valid destination, he will stop jumping.

    Mr. Kitayuta will collect the gems on the islands visited during the process. Find the maximum number of gems that he can collect.

    Input

    The first line of the input contains two space-separated integersNAndD(1? ≤?N,?D? ≤? 30000), denoting the number of the gems in the Shuseki Islands and the length of the Mr. Kitayuta's first jump, respectively.

    The nextNLines describe the location of the gems.I-Th of them (1? ≤?I? ≤?N) Contains a integerPI(D? ≤?P1? ≤?P2? ≤ ?...? ≤?PN? ≤? 30000), denoting the number of the island that containsI-Th gem.

    Output

    Print the maximum number of gems that Mr. Kitayuta can collect.

    Sample test (s) input
    4 1010212727
    Output
    3
    Input
    8 8919283645556678
    Output
    6
    Input
    13 788916171718212324242630
    Output
    4
    Note

    In the first sample, the optimal route is 0? →? 10 (+ 1 gem )? →? 19? →? 27 (+ 2 gems )? → ?...

    In the second sample, the optimal route is 0? →? 8? →? 15? →? 21? →? 28 (+ 1 gem )? →? 36 (+ 1 gem )? →? 45 (+ 1 gem )? →? 55 (+ 1 gem )? →? 66 (+ 1 gem )? →? 78 (+ 1 gem )? → ?...

    In the third sample, the optimal route is 0? →? 7? →? 13? →? 18 (+ 1 gem )? →? 24 (+ 2 gems )? →? 30 (+ 1 gem )? → ?...


    Dp is simple, but d has a large scope.

    However, it can be proved that the final d value cannot exceed 500.

    Intercept A Piece Of explanation:

    Below is the explanation from yosupo, translated by me.

    [From here]

    LetMBe the number of the islands (that is, 30001). First, let us describe a solution with time and memory complexityO(M2 ).

    We will apply Dynamic Programming. letDp[I] [J] Be the number of the gems that Mr. Kitayuta can collect after he jumps to islandI, When the length of his previous jump isJ(Let us assume that he have not collect the gems on islandI). Then, you can calculate the values of the tableDpBy the following:

    • Dp[I] [J]? =? 0, ifI? ≥?M
      (Actually these islands do not exist, but we can suppose that they exist and when Mr. Kitayuta jumps to these islands, he stops jumping)
    • Dp[I] [J]? =? (The number of the gems on islandI)? +?Max(Dp[I? +?J] [J],?Dp[I? +?J? +? 1] [J? +? 1]), ifI? M,?J? =? 1
      (He cannot perform a jump of length 0)
    • Dp[I] [J]? =? (The number of the gems on islandI)? +?Max(Dp[I? +?J? -? 1] [J? -? 1],?Dp[I? +?J] [J],?Dp[I? +?J? +? 1] [J? +? 1]), ifI? M,?J? ≥? 2

      This solution is unfeasible in terms of both time and memory. However, the following observation makes it an Accepted solution: there are only 491 valuesJThat we have to consider, which areD? -? 245 ,?D? -? 244 ,?D? -? 243 ,?...,?D? +? 244 andD? +? 245.

      Why? First, let us find the upper boundJ. Suppose Mr. Kitayuta always performs"L? +? 1 "jump (L: The length of the previous jump). Then, he will reach the end of the islands before he performs a jump of lengthD? +? 246, because
      D? +? (D? +? 1 )? +? (D? +? 2 )? + ?...? +? (D? +? (245 )? ≥? 1? +? 2? + ?...? +? 245? =? 245. (245? +? 1 )? /? 2? =? 30135?>? 30000. Thus, he will never be able to perform a jump of length.D? +? 246 or longer.

      Next, let us consider the lower boundJIn a similar way. IfD? ≤? 246, then obviously he will not be able to perform a jump of lengthD? -? 246 or shorter, because the length of a jump must be positive. Suppose Mr. Kitayuta always performs"L? -? 1 "jump, whereD? ≥? 247. Then, again he will reach the end of the islands before he performs a jump of lengthD? -? 246, because
      D? +? (D? -? 1 )? +? (D? -? 2 )? + ?...? +? (D? -? (245 )? ≥? 245? +? 244? + ?...? +? 1? =? 245. (245? +? 1 )? /? 2? =? 30135?>? 30000. Thus, he will never be able to perform a jump of length.D? -? 246 or shorter.

      Therefore, we have obtained a working solution: similar toO(M2) one, but we will only consider the valueJBetweenD? -? 245 andD? +? 245. The time and memory complexity of this solution will beO(M1.5), since the value "245" is slightly larger. <喎?http: www.bkjia.com kf ware vc " target="_blank" class="keylink"> VcD4KPHA + Signature = "normal" two dimen1_array with a offset like this: dp[i][j - offset]. The time limit is set tight in order to fail most of naive solutions with search using std: map or something, so using hash maps (unordered_map) will be risky although the complexity will be the same as the described solution.

      [End]


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                      =0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p]) #define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define MAXN (30000+10)#define MAXD (30000+10)#define M (30001)#define MP(a,b) make_pair(a,b) #define MAX_d_change (250+10)#define C (250)long long mul(long long a,long long b){return (a*b)%F;}long long add(long long a,long long b){return (a+b)%F;}long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}typedef long long ll;int n,d,a[MAXN]={0},s[MAXN]={0},f[MAXN][MAX_d_change*2]={0};int main(){//freopen("Treasure.in","r",stdin);//freopen(".out","w",stdout);cin>>n>>d;For(i,n){int p;scanf("%d",&p);a[p]++;}For(i,M) s[i]=s[i-1]+a[i];int ans=0;memset(f,-1,sizeof(f));ans=f[d][C]=a[d];Fork(i,d,M){Rep(j,2*C+1)if (f[i][j]>=0){int dis=j-C+d;if (dis>0&&i+dis<=M) {f[i+dis][j]=max(f[i+dis][j],f[i][j]+a[i+dis]);ans=max(ans,f[i+dis][j]);}if (i+dis+1<=M) {f[i+dis+1][j+1]=max(f[i+dis+1][j+1],f[i][j]+a[i+dis+1]);ans=max(ans,f[i+dis+1][j+1]);}if (dis-1>0&&i+dis-1<=M) {f[i+dis-1][j-1]=max(f[i+dis-1][j-1],f[i][j]+a[i+dis-1]);ans=max(ans,f[i+dis-1][j-1]);}}}cout<
                      







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