codeforces 488 D. Strip

D. 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.

思路：dp+线段树或平衡树优化，其中dp[i]表示以i为结尾的最少可以分成几块

/*======================================================
# Author: whai
# Last modified: 2015-12-06 10:23# Filename: d.cpp
======================================================*/
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <stack>

using namespace std;

#define LL __int64
#define PB push_back
#define P pair<int, int>
#define X first
#define Y second

const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f;

int a
;

multiset<int> myset;

int to_l
;

bool ok(int s) {
return ((*myset.rbegin()) - (*myset.begin()) <= s);
}

void predo(int n, int s, int l) {
myset.clear();
myset.insert(a[1]);
to_l[1] = 1;
int p = 1;
for(int i = 2; i <= n; ++i) {
myset.insert(a[i]);
while(p < i && !ok(s)) {
myset.erase(myset.find(a[p]));
++p;
}
to_l[i] = p;
}
//for(int i = 1; i <= n; ++i) {
//	cout<<to_l[i]<<' ';
//}
//cout<<endl;
}

int dp
;

void gao(int n, int s, int l) {
predo(n, s, l);
memset(dp, 0x3f, sizeof(dp));
myset.clear();
int p = 1;
for(int i = 1; i <= n; ++i) {
int L = to_l[i] - 1, R = i - l;
if(R > 0) myset.insert(dp[R]);
if(L > R) continue;
if(L == 0) {
dp[i] = 1;
continue;
}
while(p < L) {
myset.erase(myset.find(dp[p]));
++p;
}
int minn = (*myset.begin());
if(minn != INF)
dp[i] = minn + 1;
}
//for(int i = 1; i <= n; ++i) {
//	cout<<dp[i]<<' ';
//}
//cout<<endl;
if(dp
== INF) puts("-1");
else {
cout<<dp
<<endl;
}
}

int main() {
int n, s, l;
while(scanf("%d%d%d", &n, &s, &l) != EOF) {
for(int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
}
gao(n, s, l);
}
return 0;
}