#CodeForces - 487B#D - Strip(DP+线段树+单调队列优化)

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.

Examples

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.

题意： 将一个长度为n的数列划分成m个部分，要求每个部分含有的个数>=L，且每个部分最大值－最小值<=S，

求满足上述两个条件情况下m的最小值。即划分区间个数最小

很好的一道DP题，比较综合，把之前学过的单调队列，线段树优化全部都用上了。

定义:Dp[i]为前i个数按题目要求划分的最小段数

Dp[i] = min(Dp[j]) + 1;(其中Dp[j]为满足题意的小于i的j的Dp最小值)

借鉴HDU3530的做法：

http://blog.csdn.net/its_elaine/article/details/72677137

用单调队列按此题的方法维护最大最小值，得到j的区间，

再在线段树中搜索最小dp[j]的值

Status	Accepted
Time	61ms
Memory	6268kB
Length	1827
Lang	GNU G++ 5.1.0
Submitted	2017-06-12 15:14:35
Shared
RemoteRunId	27732877

#include
#include
#include
#include
#include
using namespace std;

const int Max = 100005;
const int INF = 0x3f3f3f3f;

struct node{
	int l, r;
	int val;
}Tr[Max << 2];

int N, S, L;
int A[Max], Qb[Max], Qm[Max];
int Dp[Max];

void getint(int & num){
	char c;	int flg = 1;	num = 0;
	while((c = getchar()) < '0' || c > '9')	if(c == '-')	flg = -1;
	while(c >= '0' && c <= '9'){	num = num * 10 + c - 48;	c = getchar();}
	num *= flg;
}

void build(int i, int l, int r){
	Tr[i].l = l, Tr[i].r = r;
	Tr[i].val = INF;
	if(l == r)	return ;
	int mid = (l + r) >> 1;
	build(i << 1, l, mid);
	build(i << 1 | 1, mid + 1, r);
}

void insert(int i, int p, int val){
	if(Tr[i].r < p || Tr[i].l > p)	return ;
	if(p == Tr[i].l && p == Tr[i].r){
		Tr[i].val = min(Tr[i].val, val);
		return ;
	}
	insert(i << 1, p, val);
	insert(i << 1 | 1, p, val);
	Tr[i].val = min(Tr[i << 1].val, Tr[i << 1 | 1].val);
}

int Query(int i, int l, int r){
	if(Tr[i].r < l || Tr[i].l > r)	return INF;
	if(l <= Tr[i].l && r >= Tr[i].r)	return Tr[i].val;
	return min(Query(i << 1, l, r), Query(i << 1 | 1, l ,r));
}

int main(){
	while(~scanf("%d%d%d", &N, &S, &L)){
		build(1, 0, N);
		insert(1, 0, 0);
		for(int i = 1; i <= N; ++ i)
			getint(A[i]);
		int lstb = 0, lstm = 0;
		int frob = 1, from = 1;
		int backb = 0, backm = 0;
		for(int i = 1; i <= N; ++ i){
			while(frob <= backb && A[Qb[backb]] < A[i])	-- backb;
			Qb[++ backb] = i;
			while(from <= backm && A[Qm[backm]] > A[i])	-- backm;
			Qm[++ backm] = i;
			while(A[Qb[frob]] - A[Qm[from]] > S){
				if(Qb[frob] < Qm[from])	lstb = Qb[frob ++];
				else lstm = Qm[from ++];
			}
			Dp[i] = Query(1, max(lstm, lstb), i - L) + 1;
			insert(1, i, Dp[i]);
		}
		if(Dp[N] >= INF)	puts("-1");
		else printf("%d\n", Dp[N]);
	}
	return 0;
}