POJ2823 Sliding Window

　　原题传送：http://poj.org/problem?id=2823

　　单调队列练手题。（不怕麻烦也可以用线段树）

　　程序运行了5000+ms，不明白几百ms的是什么算法，如果读者您知道请留个言，感激不尽。

#include <stdio.h>
#include <string.h>
#define N 1000005
int a[N];

struct node
{
    int key, val;
}e[1000005];


int main()
{
    int n, k, i, head, tail;
    while(scanf("%d%d", &n, &k) != EOF)
    {
        for(i = 1; i <= n; i ++)
            scanf("%d", &a[i]);
        
        for(head = tail = i = 1; i < k; i ++)
        {
            while(head <= tail && e[tail].val > a[i])
                tail --;
            e[++tail].key = i;
            e[tail].val = a[i];
        }
        for(i = k; i <= n; i ++)
        {
            while(head <= tail && e[head].key + k - 1 < i)
                head ++;
            while(head <= tail && e[tail].val > a[i])
                tail --;
            e[++tail].key = i;
            e[tail].val = a[i];
            printf("%d%s", e[head].val, i == n ? "" : " ");
        }
        putchar('\n');
        
        
        for(head = tail = i = 1; i < k; i ++)
        {
            while(head <= tail && e[tail].val < a[i])
                tail --;
            e[++tail].key = i;
            e[tail].val = a[i];
        }
        for(i = k; i <= n; i ++)
        {
            while(head <= tail && e[head].key + k - 1 < i)
                head ++;
            while(head <= tail && e[tail].val < a[i])
                tail --;
            e[++tail].key = i;
            e[tail].val = a[i];
            printf("%d%s", e[head].val, i == n ? "" : " ");
        }
        putchar('\n');
    }
    return 0;
}