POJ 2823 Sliding Window 单调队列

解题思路：

维护一个递增的单调队列和一个递减的单调队列，基础题。

代码：

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#define LL long long
#define FOR(i,x,y) for(int i=x;i<=y;i++)
using namespace std;
const int maxn = 1000000 + 10;
int N , M;
int A[maxn] , Q[maxn] , ID[maxn];
int head , tail;
int len1 , len2;
void init()
{
	FOR(i,1,N) scanf("%d",&A[i]);
	len1 = len2 = 0;
}
void solve_min()
{
	head = 1 , tail = 0;
	for(int i=1;i<M;i++)
	{
		while(head <= tail && Q[tail] >= A[i]) tail--;
		tail++;
		Q[tail] = A[i] ; ID[tail] = i;
	}
	FOR(i,M,N)
	{
		while(head <= tail && Q[tail] >= A[i]) tail--;
		tail++;
		Q[tail] = A[i] ; ID[tail] = i;
		while(ID[head] <= i - M) head++;
		printf("%d ",Q[head]);
	}
}
void solve_max()
{
	head = 1 , tail = 0;
	for(int i=1;i<M;i++)
	{
		while(head <= tail && Q[tail] <= A[i]) tail--;
		tail++;
		Q[tail] = A[i] ; ID[tail] = i;
	}
	FOR(i,M,N)
	{
		while(head <= tail && Q[tail] <= A[i]) tail--;
		tail++;
		Q[tail] = A[i] ; ID[tail] = i;
		while(ID[head] <= i - M) head++;
		printf("%d ",Q[head]);
	}
}
int main()
{
	while(scanf("%d%d",&N,&M)!=EOF)
	{
		init();
		solve_min();printf("\n");
		solve_max();printf("\n");
	}
	return 0;
}