codeforces 505C. Mr. Kitayuta, the Treasure Hunter (记忆化搜索)

题目大意：有30000个岛屿从左到右排列，给你一个n一个d，n代表有n个宝石分别，接下来n行表示每个宝石分别在哪个岛屿上，d代表你第一次从0开始跳跃到的位置，以后你每次可以从你的位置跳跃l-1，l，l+1的距离。

解题思路，其实以前做过一个类似的，他跳跃的步数其实很小，解设每次跳一步加以来也是（n+1）×n/2 = 30000差不多250左右，也就是说每次他最多也就会跳出来250种情况，所以，我们可以开dp[30010][500]再加一个偏移，这样记忆化搜索每个点，类似于树形dp从根节点找到一个最长的链，每次分三个叉，一个是l-1, l, l+1三个方向向下找。

C. Mr. Kitayuta, the Treasure Hunter

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

The Shuseki Islands are an archipelago of 30001 small islands in the Yutampo Sea. The islands are evenly spaced along a line, numbered from
0 to 30000 from the west to the east. These islands are known to contain many treasures. There are
n gems in the Shuseki Islands in total, and the
i-th gem is located on island
pi.

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 island
d.
After that, he will continue jumping according to the following rule. Let
l be the length of the previous jump, that is, if his previous jump was from island
prev to island cur, let
l = cur - prev. He will perform a jump of length
l - 1, l or
l + 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 when l = 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 integers
n and d (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 next n lines describe the location of the gems. The
i-th of them (1 ≤ i ≤ n) contains a integer
pi (d ≤ p1 ≤ p2 ≤ ... ≤ pn ≤ 30000),
denoting the number of the island that contains the i-th gem.

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

Sample test(s)

Input

4 10
10
21
27
27

Output

3

Input

8 8
9
19
28
36
45
55
66
78

Output

6

Input

13 7
8
8
9
1617
17
18
21
2324242630

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)
 → ...

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
///#define LL __int64#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
#define mod 1000000007
#define Read() freopen("autocomplete.in","r",stdin)
#define Write() freopen("autocomplete.out","w",stdout)
#define Cin() ios::sync_with_stdio(false)

using namespace std;

inline int read()
{
char ch;
bool flag = false;
int a = 0;
while(!((((ch = getchar()) >= '0') && (ch <= '9')) || (ch == '-')));
if(ch != '-')
{
a *= 10;
a += ch - '0';
}
else
{
flag = true;
}
while(((ch = getchar()) >= '0') && (ch <= '9'))
{
a *= 10;
a += ch - '0';
}
if(flag)
{
a = -a;
}
return a;
}
void write(int a)
{
if(a < 0)
{
putchar('-');
a = -a;
}
if(a >= 10)
{
write(a / 10);
}
putchar(a % 10 + '0');
}

const int maxn = 30010;

int dp[maxn][520];
int dx[] = {-1, 0, 1};
int Max;
int num[maxn];
int sum[maxn];
int d;

int dfs(int pos, int l)
{
int site = pos+l+d;
if(l+d < 1 || site > Max)
{
return 0;
}
if(dp[site][l+250] != -1) return dp[site][l+250];
if(l+d <= 2)
{
dp[site][l+250] = sum[site];
return dp[site][l+250];
}

if(l+d == 3)
{
dp[site][l+250] = sum[site+2]+num[site];
return dp[site][l+250];
}
int sum = 0;
for(int i = 0; i < 3; i++)
{
int xl = l+dx[i];
sum = max(sum, dfs(site, xl));
}
sum += num[site];
return dp[site][l+250] = sum;
}

int main()
{
int n;
cin >>n>>d;
memset(num, 0, sizeof(num));
memset(dp, -1, sizeof(dp));
memset(sum, 0, sizeof(sum));
int x;
Max = 0;
for(int i = 0; i < n; i++)
{
scanf("%d",&x);
Max = max(x, Max);
num[x]++;
}
for(int i = Max; i >= 1; i--) sum[i] = sum[i+1]+num[i];
x = dfs(0, 0);
cout<<x<<endl;
return 0;
}