POJ 2823 Sliding Window

题目链接：POJ 2823 Sliding Window

题目大意：

题解：
用两个双向队列\(deque\)模拟单调队列来维护区间，一个单调递增，一个单调递减，使当前区间的最大最小值分别出现在两个队列的队首。

#include <cstdio>
#include <deque>
#include <iostream>
using namespace std;

int n, k, a[1000010];

int main() {
    deque<int> maxn, minn;
    scanf("%d%d", &n, &k);
    for (int i = 0; i < n; ++i) {
        scanf("%d", &a[i]);
    }
    for (int i = 0; i < k; ++i) {  // 单调队列
        while (!minn.empty() && a[i] < minn.back()) {
            minn.pop_back();
        }
        minn.push_back(a[i]);
        while (!maxn.empty() && a[i] > maxn.back()) {
            maxn.pop_back();
        }
        maxn.push_back(a[i]);
    }
    // 窗口移动求最小
    for (int i = k; i < n; ++i) {
        cout << minn.front() << " ";
        if (minn.front() == a[i - k]) {
            minn.pop_front();
        }
        while (!minn.empty() && a[i] < minn.back()) {
            minn.pop_back();
        }
        minn.push_back(a[i]);
    }
    cout << minn.front() << " ";
    cout << endl;
    // 窗口移动求最大
    for (int i = k; i < n; ++i) {
        cout << maxn.front() << " ";
        if (maxn.front() == a[i - k]) {
            maxn.pop_front();
        }
        while (!maxn.empty() && a[i] > maxn.back()) {
            maxn.pop_back();
        }
        maxn.push_back(a[i]);
    }
    cout << maxn.front() << " ";
    cout << endl;
    return 0;
}

标签：minn,int,back,pop,队列,Window,maxn,POJ,Sliding
来源： https://www.cnblogs.com/IzumiSagiri/p/15860609.html