Strip CodeForces - 487B (单调队列)

题面:

Alexandra has a paper strip with n numbers on it. Let's call them a_i 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.

一个显然的思路是ST表求一下最值, 由于最值单调性可以双指针处理出以每个数为右端点时, 左端点的最小值, 然后再dp就行了, 但这样复杂度是$O(nlogn)$的. 若用单调队列处理的话可以达到$O(n)$的, 单调队列还是不太会写啊, 写了1个多小时才A

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <queue>
#define REP(i,a,n) for(int i=a;i<=n;++i)
using namespace std;

const int N = 1e6+10, INF = 0x3f3f3f3f;
int n, s, l;
int a[N], dp[N], L[N];

int main() {
	scanf("%d%d%d", &n, &s, &l);
	REP(i,1,n) scanf("%d", a+i);
	deque<int> q;
	int pos = 1;
	REP(i,1,n) {
		while (q.size()&&a[i]-a[q.front()]>s) pos=q.front()+1,q.pop_front();
		L[i] = pos;
		while (q.size()&&a[i]<a[q.back()]) q.pop_back();
		q.push_back(i);
	}
	pos = 1, q.clear();
	REP(i,1,n) {
		while (q.size()&&a[q.front()]-a[i]>s) pos=q.front()+1,q.pop_front();
		L[i] = max(L[i], pos);
		while (q.size()&&a[i]>a[q.back()]) q.pop_back();
		q.push_back(i);
	}
	REP(i,1,n) dp[i]=INF;
	q.clear();
	q.push_back(0);
	REP(i,1,n) {
		while (q.size()&&q.front()<L[i]-1) q.pop_front();
		if (q.size()&&q.front()<=i-l) dp[i]=dp[q.front()]+1;
		while (q.size()&&dp[i]<dp[q.back()]) q.pop_back();
		q.push_back(i);
	}
	printf("%d\n", dp[n]>=INF?-1:dp[n]);
}