POJ 2823 Sliding Window(单调队列)

单调队列典型题

An array of size n ≤ 10 ⁶ is given to you. There is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves rightwards by one position. Following is an example:
The array is [1 3 -1 -3 5 3 6 7], and k is 3.

Window position	Minimum value	Maximum value
[1  3  -1] -3  5  3  6  7	-1	3
1 [3  -1  -3] 5  3  6  7	-3	3
1  3 [-1  -3  5] 3  6  7	-3	5
1  3  -1 [-3  5  3] 6  7	-3	5
1  3  -1  -3 [5  3  6] 7	3	6
1  3  -1  -3  5 [3  6  7]	3	7

Your task is to determine the maximum and minimum values in the sliding window at each position.

#include<cstdio>  
#include<cstring>  
#include<cmath>  
#include<cstdlib>  
#include<iostream>  
#include<algorithm>  
#include<vector>  
#include<map>  
#include<queue>  
#include<stack> 
#include<string>
#include<map> 
#include<set>
#define eps 1e-6 
#define LL long long  
using namespace std;  

const int maxn = 2000000;
//const int INF = 0x3f3f3f3f;
int n, k;
int a[maxn];
//单调队列
int qmin[maxn], vmin[maxn], hmin = 1, tmin = 0; 
void Min(int a, int i) {  //第i个元素a入队 
	while(hmin<=tmin && vmin[hmin] <= i-k) hmin++;  //超范围队首出队 
	//while(hmin<=tmin && qmin[tmin]>=a) tmin--; //不符合要求队尾出列 
	int l = hmin, r = tmin;
	while(l <= r) {
		int m = l+(r-l)/2;
		if(qmin[m] >= a) r = m - 1;
		else l = m + 1; 
	}
	tmin = ++r;
	qmin[tmin] = a;
	vmin[tmin] = i;
}
int qmax[maxn], vmax[maxn], hmax = 1, tmax = 0; 
void Max(int a, int i) {  //第i个元素a入队 
	while(hmax<=tmax && vmax[hmax] <= i-k) hmax++;  //超范围队首出队 
	//while(hmax<=tmax && qmax[tmax]<=a) tmax--; //不符合要求队尾出列 
	int l = hmax, r = tmax;
	while(l <= r) {
		int m = l+(r-l)/2;
		if(qmax[m] <= a) r = m - 1;
		else l = m + 1; 
	}
	tmax = ++r;
	qmax[tmax] = a;
	vmax[tmax] = i;
}
int ansMax[maxn], ansMin[maxn];
int main() {
//	freopen("input.txt", "r", stdin);
	while(scanf("%d%d", &n, &k) == 2) {
		hmin = 1, tmin = 0;
		hmax = 1, tmax = 0;
		for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
		for(int i = 1; i < k; i++) {
			Min(a[i], i);
			Max(a[i], i);
		}
		for(int i = k; i <= n; i++) {
			Min(a[i], i);
			ansMin[i-k] = qmin[hmin];
			Max(a[i], i);
			ansMax[i-k] = qmax[hmax];
		}
		for(int i = 0; i <= n-k; i++) 
			if(i != n-k) printf("%d ", ansMin[i]);
			else printf("%d\n", ansMin[i]);
		for(int i = 0; i <= n-k; i++) 
			if(i != n-k) printf("%d ", ansMax[i]);
			else printf("%d\n", ansMax[i]);
	}
	return 0;
}