Evolution Game

In the fantasy world of ICPC there are magical beasts. As they grow, these beasts can change form, and every time they do they become more powerful. A beast cannot change form completely arbitrarily though. In each form a beast has n eyes and k horns, and these affect the changes it can make.  
 
A beast can only change to a form with more horns than it currently has. 
A beast can only change to a form that has a difference of at most w eyes. So, if the beast currently has n eyes it can change to a form with eyes in range [n - w, n + w]. 
  
A beast has one form for every number of eyes between 1 and N, and these forms will also have an associated number of horns. A beast can be born in any form. The question is, how powerful can one of these beasts become? In other words, how many times can a beast change form before it runs out of possibilities? 

输入

The first line contains two integers, N and w, that indicate, respectively, the maximum eye number, and the maximum eye difference allowed in a change (1 ≤ N ≤ 5000; 0 ≤ w ≤ N).  
The next line contains N integers which represent the number of horns in each form. I.e. the ith number, h(i), is the number of horns the form with i eyes has (1 ≤ h(i) ≤ 1 000 000). 

输出

For each test case, display one line containing the maximum possible number of changes. 

样例输入

复制样例数据

5 5
5 3 2 1 4

样例输出

4

提示

Start with 1 horn and 4 eyes, and it can change 4 times: (1 horn 4 eyes) -> (2 horns 3 eyes) -> (3 horns 2 eyes) -> (4 horns 5 eyes) -> (5 horns 1 eye). 

ps:dp,从后往前遍历，更新每个状态，寻找最大值

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
const int inf=1e9,N=5555;
int a[N];
int dp[N];
int main()
{
    int n,w,x=0;
    int ans=0;
    cin>>n>>w;
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        if(a[i]>a[x]) x=i;
    }
    int m=n;
    while(m--){
        for(int i=1;i<=n;i++){
            if(a[i]<a[x]&&abs(i-x)<=w){
                dp[i]=max(dp[i],dp[x]+1);
                ans=max(ans,dp[i]);
            }
        }
        a[x]=inf;
        x=0;
        for(int i=1;i<=n;i++){
            if(a[i]!=inf&&a[i]>a[x]) x=i;
        }
    }
    cout<<ans<<endl;
    return 0;
}