栈和队列专题

此处为栈和队列的专题

【练习】栈和队列 - Virtual Judge (vjudge.net)

A - Train Problem I

B - 简单计算器

C - Team Queue

D - 愚人节的礼物

E - ACboy needs your help again!

F - 看病要排队

G - Windows Message Queue

题目详情 - 滑动窗口 /【模板】单调队列 - 可达编程 (kedacun.ac)

思路：模板题，就是自己在本子上模拟即可（或者去B站搜搜视频）

这题刚好取两个极限，一个最大一个最小

#include<iostream>
const int N = 1e6+10;
int a[N],q[N];
int n,m;
using namespace std;
int main()
{
    cin>>n>>m;
    for(int i = 1;i <= n;i++)
    {
        cin>>a[i];
    }
    int h = 0,t = -1;
    for(int i = 1;i <= n;i++)
    {
        while(h <= t && a[i] <= a[q[t]]) t--;
        q[++t] = i;
        if(h <= t && q[h] < i-m+1)h++;
        if(i > m-1)cout<<a[q[h]]<<" ";
    }
cout<<endl;
h = 0,t = -1;
for(int i = 1;i <= n;i++)
    {
       while(h <= t && a[i] >= a[q[t]])t--;
       q[++t] = i;
       if(h <= t && q[h] < i-m+1)h++;
       if(i > m-1)cout<<a[q[h]]<<" ";;
    }


    return 0;
}

  

题目详情 - [USACO12MAR] Flowerpot S - 可达编程 (kedacun.ac) 

思路：一个单调队列即可 

#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1e6+10;
int q[N];
int ans = 0x3f3f3f;
int n,m;
struct node
{
    int x,y;
}a[N];

bool cmp(node x,node y)
{
    return x.x < y.x;
}
int main()
{
    int ma = 0,mi = 0x3f3f3f;
    cin>>n>>m;
    for(int i = 1;i <= n;i++)
    {
        cin>>a[i].x>>a[i].y;
        ma = max(ma,a[i].y);
        mi = min(mi,a[i].y);
    }
    sort(a+1,a+n+1,cmp);
    int h = 1,t = 0;
    for(int i = 1;i <= n;i++)
    {
        while(h <= t && a[i].y <= a[q[t]].y) t--;
        q[++t] = i;
        while(h <= t && a[q[t]].y - a[q[h]].y >= m)ans = min(ans,a[q[t]].x - a[q[h]].x),h++;
    }
    if(ans == 0x3f3f3f)cout<<"-1"<<endl;
    else cout<<ans<<endl;

    return 0;
}

 题目详情 - [SCOI2009] 生日礼物 - 可达编程 (kedacun.ac)

思路：继续单调队列模拟即可 

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
const int N = 1e6+10;
struct zz
{
    int a,b;
}d[N],q[N];
int a[N];
int cnt;//用来记录每一次的珠子颜色
int minn = 0x3f3f3f3f,x,num;
int n,k;
inline bool cmp(zz x,zz y)
{
    return x.a < y.a;
}
int main()
{
    cin>>n>>k;
    for(int i = 1;i <= k;i++)
    {
        cin>>x;
        for(int j = 1;j <= x;j++)
        {
            cin>>d[++num].a;
            d[num].b = i;
        }
    }
    int h = 1,t = 0;
    sort(d+1,d+num+1,cmp);
    for(int i = 1;i <= n;i++)
    {
        q[++t] = d[i];
        a[d[i].b]++;
        if(a[d[i].b] == 1)cnt++;

        while(cnt == k)
        {
            minn = min(minn,q[t].a - q[h].a);
            a[q[h].b]--;
            if(a[q[h].b] == 0)cnt--;
            h++;
        }
    }
    cout<<minn<<endl;

    return 0;
}