「CF506A」Mr. Kitayuta, the Treasure Hunter

2015年1月19日3,3630

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 p_i.

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 p_i (d ≤ p₁ ≤ p₂ ≤ … ≤ p_n ≤ 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

output

3

input

8 8

output

6

input

13 7

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) → …

题解

3W*3W的dp是一种显然的做法，f[i][j]表示到第i个位置，上次跳跃j获得的最大价值

仔细思考后发现，不合法的情况很多，即j并不用从1枚举到30000

j的变化量只有根号级别

于是第二维缩减到根号n

#include<iostream>
#include<set>
#include<map>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<ctime>
#include<vector>
#include<queue>
#include<algorithm>
#include<cmath>
#define T 600
#define inf 2000000000
#define pa pair<int,int>
#define ll long long 
using namespace std;
ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,d,ans;
int p[30005];
int f[30005][605];
int main()
{
    memset(f,-1,sizeof(f));
    n=read();d=read();
    for(int i=1;i<=n;i++)
    {
        int x=read();
        p[x]++;
    }
    f[d][T/2]=0;
    for(int i=d;i<=30000;i++)
    {
        for(int j=1;j<=T;j++)
        {
            int t=j-T/2+d;
            if(i-t>=0&&t>0)
            {
                if(f[i-t][j-1]!=-1)f[i][j]=max(f[i][j],f[i-t][j-1]);
                if(f[i-t][j]!=-1)f[i][j]=max(f[i][j],f[i-t][j]);
                if(f[i-t][j+1]!=-1)f[i][j]=max(f[i][j],f[i-t][j+1]);
                if(f[i][j]!=-1)f[i][j]+=p[i];
                ans=max(f[i][j],ans);
            }
        }
    }
    printf("%d\n",ans);
    return 0;
}

#include<iostream>

#include<set>

#include<map>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<ctime>

#include<vector>

#include<queue>

#include<algorithm>

#include<cmath>

#define T 600

#define inf 2000000000

#define pa pair<int,int>

#define ll long long

using namespace std;

ll read()

{

ll x=0,f=1;char ch=getchar();

while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

return x*f;

}

int n,d,ans;

int p[30005];

int f[30005][605];

int main()

{

memset(f,-1,sizeof(f));

n=read();d=read();

for(int i=1;i<=n;i++)

{

int x=read();

p[x]++;

}

f[d][T/2]=0;

for(int i=d;i<=30000;i++)

{

for(int j=1;j<=T;j++)

{

int t=j-T/2+d;

if(i-t>=0&&t>0)

{

if(f[i-t][j-1]!=-1)f[i][j]=max(f[i][j],f[i-t][j-1]);

if(f[i-t][j]!=-1)f[i][j]=max(f[i][j],f[i-t][j]);

if(f[i-t][j+1]!=-1)f[i][j]=max(f[i][j],f[i-t][j+1]);

if(f[i][j]!=-1)f[i][j]+=p[i];

ans=max(f[i][j],ans);

}

printf("%d\n",ans);

return 0;

}

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「CF506A」Mr. Kitayuta, the Treasure Hunter

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