C. Mr. Kitayuta, the Treasure Hunter
time limit per test
memory limit per test
input
output
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.
0. With his great jumping ability, he will repeatedly perform jumps between islands to the east according to the following process:
- 0 to islandd.
- lbe 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
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.
n lines describe the location of the gems. The i-th of them (1 ≤ i ≤ n) contains a integer pi (d ≤ p1 ≤ p2 ≤ ... ≤ pn), 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 16 17 17 18 21 23 24 24 26 30
output
4
Note
→ 10 (+1 gem) → 19 → 27 (+2 gems) → ...
→ 8 → 15 → 21 → 28 (+1 gem) → 36 (+1 gem) → 45 (+1 gem) → 55 (+1 gem) → 66 (+1 gem) → 78 (+1 gem) → ...
→ 7 → 13 → 18 (+1 gem) → 24 (+2 gems) → 30 (+1 gem) → ...
这题直接dp很简单，但是d的范围很大
但是可以证明最终用到的d的取值最多不超过500
截取一段explanation:
Below is the explanation from yosupo, translated by me.
[From here]
m be the number of the islands (that is, 30001). First, let us describe a solution with time and memory complexity of O(m2).
dp[i][j] be the number of the gems that Mr. Kitayuta can collect after he jumps to island i, when the length of his previous jump is j (let us assume that he have not collect the gems on island i). Then, you can calculate the values of the table dp
- 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
491 values of j that we have to consider, which are d - 245, d - 244, d - 243, ..., d + 244 and d + 245.
j. Suppose Mr. Kitayuta always performs the "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 length d + 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
j in a similar way. If d ≤ 246, then obviously he will not be able to perform a jump of length d - 246 or shorter, because the length of a jump must be positive. Suppose Mr. Kitayuta always performs the "l - 1" jump, where d ≥ 247. Then, again he will reach the end of the islands before he performs a jump of length d - 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
O(m2) one, but we will only consider the value of j between d - 245 andd + 245. The time and memory complexity of this solution will be O(m1.5), since the value "245" is slightly larger than
.
This solution can be implemented by, for example, using a "normal" two dimensional 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]
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<map>
#include<vector>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=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<<ans<<endl;
return 0;
}
