关于P1886 滑动窗口 /【模板】单调队列

这题快给我写吐了

单调队列模板题，我想试试用deque写

然后

#include <iostream>
#include <deque>
using namespace std;
deque<int> p;int l, r, n, k, a[1000001];
struct min_q
{
    deque<int> d;
    int front(){return d.front();}
    void pop()
    {
        if(p.front() < l)
            d.pop_front(), p.pop_front();
    }
    void push(int x)
    {
        if(!d.empty())
            while(x < d.back())
            {
                d.pop_back();p.pop_back();
                if(d.empty()) break;
            }
        d.push_back(x);
    }
}x;
struct max_q
{
    deque<int> d;
    int front(){return d.front();}
    void pop()
    {
        if(p.front() < l)
            d.pop_front(), p.pop_front();
    }
    void push(int x)
    {
        if(!d.empty())
            while(x > d.back())
            {
                d.pop_back();p.pop_back();
                if(d.empty()) break;
            }
        d.push_back(x);
    }
}y;
int main()
{
    cin >> n >> k;r = k;
    for(int i = 0;i < n;++i)
        cin >> a[i];
    for(int i = 0;i < k;++i)
        x.push(a[i]), p.push_back(i);
    cout << x.front() << " ";
    for(int i = k;i < n;++i)
    {
        ++l;++r;x.pop();
        x.push(a[i]);p.push_back(i);
        cout << x.front() << " ";
    }
    cout << endl;
    l = 0;r = k;
    while(!p.empty()) p.pop_back();
    for(int i = 0;i < k;++i)
        y.push(a[i]), p.push_back(i);
    cout << y.front() << " ";
    for(int i = k;i < n;++i)
    {
        ++l;++r;y.pop();
        y.push(a[i]);p.push_back(i);
        cout << y.front() << " ";
    }
    return 0;
}

72行超短代码

这个故事告诉我们$STL$会增加代码长度

压了压行之后

#include <iostream>
#include <deque>
using namespace std;
deque<int> p;int l, r, n, k, a[1000001];
bool c1(int a, int b) {return a < b;}
bool c2(int a, int b) {return a > b;}
struct q
{
    deque<int> d;bool (*c)(int a, int b);
    q(bool (*b)(int a, int b)) {c = b;}
    int front(){return d.front();}
    void pop()
    {
        if(p.front() < l)
            d.pop_front(), p.pop_front();
    }
    void push(int x)
    {
        if(!d.empty())
            while((*c)(x, d.back()))
            {
                d.pop_back();p.pop_back();
                if(d.empty()) break;
            }
        d.push_back(x);
    }
}x(c1), y(c2);
void work(q z)
{
    for(int i = 0;i < k;++i)
        z.push(a[i]), p.push_back(i);
    cout << z.front() << " ";
    for(int i = k;i < n;++i)
    {
        ++l;++r;z.pop();
        z.push(a[i]);p.push_back(i);
        cout << z.front() << " ";
    }
}
int main()
{
    cin >> n >> k;r = k;
    for(int i = 0;i < n;++i)
        cin >> a[i];
    work(x);
    cout << endl;
    l = 0;r = k;
    while(!p.empty()) p.pop_back();
    work(y);
    return 0;
}

不管怎么说反正这道题是A了