给你一个长度为 N 的数组，一个长为 K 的滑动的窗体从最左移至最右端，你只能见到窗口的 K 个整数，每次窗体向右移动一位，如下表：

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你的任务是找出窗口在各位置时的最大值和最小值。

输入的第 1 行是两个整数 n，k，第 2 行为长度为 n 的数组（即有 n 个整数）。

输出 2 行，第 1 行是每个位置的最小值，第 2 行是每个位置的最大值。

输入　 [复制]

8 3 
1 3 -1 -3 5 3 6 7

输出

-1 -3 -3 -3 3 3 
3 3 5 5 6 7

【数据范围】

对于 20% 的数据：n<=500；
对于 50% 的数据：n<=100000；
对于 100% 的数据：n<=1000000；

题目分析

入门级别单调队列。分别维护一个单调递增和单调递减的队列，每次添加新数并维护单调性，然后删除队首的在区间外面的数，最后队首就是答案。

code

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<cstring>

#include<string>

#include<algorithm>

using namespace std;

const int N = , oo = 0x3f3f3f3f;

typedef pair<int, int> P;

int data[N], head, tail, n, k;

P que[N];

inline int read(){

    int i = , f = ; char ch = getchar();

    for(; (ch < '' || ch > '') && ch != '-'; ch = getchar());

    if(ch == '-') f = -, ch = getchar();

    for(; ch >= '' && ch <= ''; ch = getchar())

        i = (i << ) + (i << ) + (ch - '');

    return i * f;

}

inline void wr(int x){

    if(x < ) putchar('-'), x = -x;

    if(x > ) wr(x / );

    putchar(x %  + '');

}

int main(){

    n = read(), k = read();

    for(int i =  ; i <= n; i++) data[i] = read();

    tail = head = ;

    que[head].second = oo;

    for(int i = ; i <= n; i++){

        if(data[i] < que[head].second)

            que[head = tail = ] = P(i, data[i]);

        else{

            while(head < tail && que[tail].second > data[i]) tail--;

            que[++tail] = P(i, data[i]);

        }

        while(que[head].first <= i - k) head++;

        if(i >= k)

            wr(que[head].second), putchar(' ');

//        cout<<endl<<"!!";for(int j = head; j <= tail; j++) cout<<"("<<que[j].first<<") "<<que[j].second<<" ";cout<<endl;

    }

    putchar('\n');

    tail = head = ;

    que[head].second = -oo;

    for(int i = ; i <= n; i++){

        if(data[i] > que[head].second)

            que[head = tail = ] = P(i, data[i]);

        else{

            while(head < tail && que[tail].second < data[i]) tail--;

            que[++tail] = P(i, data[i]);

        }

        while(que[head].first <= i - k) head++;

        if(i >= k)

            wr(que[head].second), putchar(' ');

    //        cout<<endl<<"!!";for(int j = head; j <= tail; j++) cout<<que[j].first<<" "<<que[j].second<<" ";cout<<endl;

    }

    return ;

}