Codeforces Round #278 (Div. 2) D. Strip 线段树优化dp-优快云博客

本文介绍了一个关于纸条分割的问题：给定一条包含数字的纸条，如何将其分割成满足特定条件的最少段数。每段至少包含l个数字，且最大与最小数字之差不超过s。文章提供了一种使用线段树进行区间查询的方法来解决此问题。

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Alexandra has a paper strip with n numbers on it. Let's call them ai from left to right.

Now Alexandra wants to split it into some pieces (possibly 1). For each piece of strip, it must satisfy:

Each piece should contain at least l numbers.
The difference between the maximal and the minimal number on the piece should be at most s.

Please help Alexandra to find the minimal number of pieces meeting the condition above.

Input

The first line contains three space-separated integers n, s, l (1 ≤ n ≤ 105, 0 ≤ s ≤ 109, 1 ≤ l ≤ 105).

The second line contains n integers ai separated by spaces ( - 109 ≤ ai ≤ 109).

Output

Output the minimal number of strip pieces.

If there are no ways to split the strip, output -1.

Examples

input

7 2 2
1 3 1 2 4 1 2

output

input

7 2 2
1 100 1 100 1 100 1

output

-1

Note

For the first sample, we can split the strip into 3 pieces: [1, 3, 1], [2, 4], [1, 2].

For the second sample, we can't let 1 and 100 be on the same piece, so no solution exists.

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e5+10,M=1e6+10,inf=2e9;
const ll INF=1e18+10,mod=2147493647;
int a[N];
struct linetree
{
    int maxx[N<<2],minn[N<<2];
    void pushup(int pos)
    {
        maxx[pos]=max(maxx[pos<<1],maxx[pos<<1|1]);
        minn[pos]=min(minn[pos<<1],minn[pos<<1|1]);
    }
    void build(int l,int r,int pos)
    {
        if(l==r)
        {
            maxx[pos]=a[l];
            minn[pos]=a[l];
            return;
        }
        int mid=(l+r)>>1;
        build(l,mid,pos<<1);
        build(mid+1,r,pos<<1|1);
        pushup(pos);
    }
    void update(int p,int c,int l,int r,int pos)
    {
        if(l==r)
        {
            maxx[pos]=c;
            minn[pos]=c;
            return;
        }
        int mid=(l+r)>>1;
        if(p<=mid)
            update(p,c,l,mid,pos<<1);
        else
            update(p,c,mid+1,r,pos<<1|1);
        pushup(pos);
    }
    int query(int L,int R,int l,int r,int pos,int flag)
    {
        if(L<=l&&r<=R)
        {
            if(flag)
                return maxx[pos];
            else
                return minn[pos];
        }
        int mid=(l+r)>>1;
        int ans=-inf;
        if(!flag)
            ans=inf;
        if(L<=mid)
            if(flag)
                ans=max(ans,query(L,R,l,mid,pos<<1,flag));
            else
                ans=min(ans,query(L,R,l,mid,pos<<1,flag));
        if(R>mid)
            if(flag)
                ans=max(ans,query(L,R,mid+1,r,pos<<1|1,flag));
            else
                ans=min(ans,query(L,R,mid+1,r,pos<<1|1,flag));
        return ans;
    }
};
linetree tree,dp;
int main()
{
    int n,l,s;
    scanf("%d%d%d",&n,&s,&l);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    tree.build(1,n+1,1);
    dp.build(1,n+1,1);
    dp.update(1,0,1,n+1,1);
    for(int i=1;i<=n;i++)
    {
        int st=0,en=i-l,ans=-1;
        while(st<=en)
        {
            int mid=(st+en)>>1;
            if(tree.query(mid+1,i,1,n+1,1,1)-tree.query(mid+1,i,1,n+1,1,0)<=s)
            {
                ans=mid;
                en=mid-1;
            }
            else
                st=mid+1;
        }
        //cout<<ans<<endl;
        if(ans==-1||ans+1>i-l+1)
            dp.update(i+1,inf,1,n+1,1);
        else
        {
            int minn=dp.query(ans+1,i-l+1,1,n+1,1,0);
            //cout<<ans+1<<" "<<i-l+1<<" "<<i<<" "<<minn<<endl;
            dp.update(i+1,minn+1,1,n+1,1);
        }

    }
    if(dp.query(n+1,n+1,1,n+1,1,1)>=inf)
        printf("-1");
    else
        printf("%d\n",dp.query(n+1,n+1,1,n+1,1,1));
    return 0;
}
///  dp[i]=min(dp[mid-1]-dp[i-l])+1