【poj2823】Sliding Window

Description 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.  Input The input consists of two lines. The first line contains two integers  n and  k which are the lengths of the array and the sliding window. There are  n integers in the second line.  Output There are two lines in the output. The first line gives the minimum values in the window at each position, from left to right, respectively. The second line gives the maximum values.  Sample Input 8 3 1 3 -1 -3 5 3 6 7 Sample Output -1 -3 -3 -3 3 3 3 3 5 5 6 7 Source POJ Monthly--2006.04.28, Ikki

【题解】

单调队列裸题 详见本博客单调队列学习笔记
【代码】

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int a[1000005];
int queue[1000005];
int n,k,head,tail;
inline void push1(int x,int i){
	if (!head&&!tail||head>=tail){
		queue[++tail]=x; return;
	}
	while (queue[head+1]<i&&head<tail) head++;
	while ((a[queue[tail]]>a[x])&&head<tail) tail--;
	queue[++tail]=x;
}
inline void push2(int x,int i){
	if (!head&&!tail||head>=tail){
		queue[++tail]=x; return;
	}
	while (queue[head+1]<i&&head<tail) head++;
	while ((a[queue[tail]]<a[x])&&head<tail) tail--;
	queue[++tail]=x;
}
int main(){
	scanf("%d%d",&n,&k);
	for (int i=1;i<=n;++i) scanf("%d",&a[i]);
	for (int i=1;i<=k;++i) push1(i,1);
	printf("%d ",a[queue[head+1]]);
	for (int i=k+1;i<=n;++i){
		push1(i,i-k+1);
		printf("%d",a[queue[head+1]]);
		if (i==n) printf("\n"); else printf(" ");
	}
	head=tail=0; memset(queue,0,sizeof(queue));
	for (int i=1;i<=k;++i) push2(i,1);
	printf("%d ",a[queue[head+1]]);
	for (int i=k+1;i<=n;++i){
		push2(i,i-k+1);
		printf("%d",a[queue[head+1]]);
		if (i==n)printf("\n"); else printf(" ");
	}
}