Codeforces Round #286 (Div. 2) C dp

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C. Mr. Kitayuta, the Treasure Hunter

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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.

Examples
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
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)  → ...

题意：

有30001个岛屿，部分岛屿上有宝藏，一个人每次能跳d|d+1|d-1的距离个岛屿，初始值d给定。

问最大能获得多少宝藏

思路：

dp[i][j]表示到达岛屿i且步数为j的最大宝藏值，但是i,j都可以是30001，会超内存

考虑到d的变化每次都只能是+-1，设到达终点需要跳num次

那么有：d+d+1+d+2+...+d+num-1=num*d+(num-1)*num/2 <=30001

=> d<=250  但是d可以是减少或者增加250。那么开500的数组 [0,250]存-1的 [250,500] 存+1的

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<string>#include<vector>#include <ctime>#include<queue>#include<set>#include<map>#include<stack>#include<iomanip>#include<cmath>#include<bitset>#define mst(ss,b) memset((ss),(b),sizeof(ss))///#pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef long double ld;#define INF (1ll<<60)-1#define Max 1e9using namespace std;int dp[30010][550]; /// 当前在第i个岛屿，跳的步数为j的最大宝藏int n,d,pos;int a[30010];int main(){    scanf("%d%d",&n,&d);    for(int i=1;i<=n;i++){        scanf("%d",&pos);        a[pos]++;    }    mst(dp,-1);    dp[d][250]=0;    int ans=0;    for(int i=d;i<=30001;i++){        for(int j=0;j<=500;j++){            if(dp[i][j]==-1) continue;            dp[i][j]+=a[i];            ans=max(ans,dp[i][j]);            for(int k=j-1;k<=j+1;k++){                int next=i+k-250+d;                if(next<=i || k<=0 || next>30001) continue;                dp[next][k]=max(dp[next][k],dp[i][j]);            }        }    }    cout<<ans<<endl;    return 0;}