Codeforces 487B. Strip DP+线段树+二分

dp[ i ]表示到第i个位置最少要分多少下, dp[ i ] = min ( dp [ i ] , dp [ j ] + 1 ) j 在合适的范围内 (  满足长度和最值差 )

对整个数组建立线段树维护最大值和最小值这样就可在nlogn的时间里求出某一段的最值差,这个范围是满足单调性的,所以对于每个i可以二分出j的最小值 . 

对每个dp[i]建立线段树,可以在nlogn时间内求出最小的j.

所以总时间复杂度n^2logn 

B. Strip

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

input

7 2 2
1 3 1 2 4 1 2

output

3

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.

/* ***********************************************
Author        :CKboss
Created Time  :2015年03月11日 星期三 23时20分17秒
File Name     :CF487B.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int maxn = 100100;
const int INF = 1e6;

int n,l,s;
int a[maxn];

int maxnum[maxn<<2],minnum[maxn<<2];

/// seg tree

#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef pair<int,int> pII;

void push_up(int rt)
{
	maxnum[rt]=max(maxnum[rt<<1],maxnum[rt<<1|1]);
	minnum[rt]=min(minnum[rt<<1],minnum[rt<<1|1]);
}

void build(int l,int r,int rt)
{
	if(l==r)
	{
		minnum[rt]=maxnum[rt]=a[l];
		return ;
	}
	int m=(l+r)/2;
	build(lson); build(rson);
	push_up(rt);
}

pII query(int L,int R,int l,int r,int rt)
{
	if(L<=l&&r<=R)
	{
		return make_pair(maxnum[rt],minnum[rt]);
	}

	int m=(l+r)/2;
	if(R<=m) return query(L,R,lson);
	else if(L>m) return query(L,R,rson);
	else
	{
		pII leftp=query(L,R,lson);
		pII rightp=query(L,R,rson);
		return make_pair( max( leftp.first,rightp.first) 
				, min( leftp.second, rightp.second ) ); 
	}
}

int dp[maxn];
/// seg tree 2
int tree2[maxn<<2];

void push_up2(int rt)
{
	tree2[rt]=min(tree2[rt<<1],tree2[rt<<1|1]);
}

void build2()
{
	memset(tree2,63,sizeof(tree2));
}

void insert2(int val,int pos,int l,int r,int rt)
{
	if(pos==l&&pos==r)
	{
		tree2[rt]=val;
		return ;
	}
	int m=(l+r)/2;
	if(pos<=m) insert2(val,pos,lson);
	else insert2(val,pos,rson);
	push_up2(rt);
}

int query2(int L,int R,int l,int r,int rt)
{
	if(L<=l&&r<=R)
	{
		return tree2[rt];
	}
	int m=(l+r)/2;
	if(R<=m) return query2(L,R,lson);
	else if(L>m) return query2(L,R,rson);
	else
	{
		int one = query2(L,R,lson);
		int two = query2(L,R,rson);
		return min(one,two);
	}
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	scanf("%d%d%d",&n,&s,&l);
	for(int i=1;i<=n;i++) 
	{
		dp[i]=INF;
		scanf("%d",a+i);
	}

	build(1,n,1);
	dp[0]=0; dp[1]=1;
	if(l>1) dp[1]=INF;

	build2();
	insert2(0,0,0,n,1);
	insert2(1,1,0,n,1);

	for(int i=2;i<=n;i++)
	{
		/// binary search to find left bound
		int low=1,high=i-l+1,leftb=-1;
		while(low<=high)
		{
			int mid=(low+high)/2;
			pII temp = query(mid,i,1,n,1);
			if(temp.first-temp.second<=s)
			{
				leftb=mid; high=mid-1;
			}
			else  low=mid+1;
		}
		if(leftb==-1) dp[i]=INF;
		else
		{
			int mmm = query2(leftb-1,i-l,0,n,1);
			dp[i]=mmm+1;
		}
		insert2(dp[i],i,0,n,1);
	}
	if(dp[n]>=INF) dp[n]=-1;
	printf("%d\n",dp[n]);

    return 0;
}