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单调队列(滑动窗口) · GitBook

• Introduction

• 并查集

• 差分

• 尺取法双指针法

• 大数

• 单调队列(滑动窗口)

• 单调栈

• 动态规划

• 二分

• 分治

• 哈希表

• 链表

• 模拟

• 排序

• 前缀和

• 三指针法

• 设计

• 树

• 数论

• 数学

• 数组

• 搜索回溯&枚举

• 拓扑排序

• 贪心

• 图

• 位运算

• 优先队列

• 栈

• 字符串

• 最小生成树

• LeetCode未分类

• Manacher

• Offer

• Ordered Map

• Published with GitBook

单调队列(滑动窗口)

单调队列+滑动窗口

[TOC]

• 模板

• 

• 代码：

  #include<iostream> //https://www.luogu.com.cn/problem/P1886

  using namespace std; //双向单调队列

  const int maxn=1000005;

  int a[maxn],q[maxn],p[maxn],head=1,tail=0; //head,tail:相对的头和尾

  int n,k;

  void monotone_min(){ //单调递增(q[head]最小)

  head=1,tail=0; //head=1,tail=0

  for(int i=1;i<=n;i++){

  while(head<=tail&&q[tail]>a[i]){ //q[tail]>a[i](不满足的筛选掉)

  tail--; //队尾出队

  }

  q[++tail]=a[i]; //入队

  p[tail]=i; //记录当前队尾索引

  if(p[head]<=i-k){ //超出滑动窗口(例:[-3,3,6])

  head++; //队首出队

  }

  if(i>=k){ //满足滑动窗口条件(从第一个窗口往后,每移动一位,输出一个元素)

  cout<<q[head]<<" "; //输出最小元素(队首)

  }

  }

  cout<<endl;

  }

  void monotone_max(){ //单调递减(q[head]最大)

  head=1,tail=0;

  for(int i=1;i<=n;i++){

  while(head<=tail&&q[tail]<a[i]){ //q[tail]<a[i]

  tail--;

  }

  q[++tail]=a[i];

  p[tail]=i;

  if(p[head]<=i-k){

  head++;

  }

  if(i>=k){

  cout<<q[head]<<" ";

  }

  }

  cout<<endl;

  }

  int main(){

  ios::sync_with_stdio(0);

  cin.tie(0);

  cin>>n>>k;

  for(int i=1;i<=n;i++){

  cin>>a[i];

  }

  monotone_min();

  monotone_max();

  return 0;

  }

  /*

  8 3

  1 3 -1 -3 5 3 6 7

  -1 -3 -3 -3 3 3

  3 3 5 5 6 7

  */

题

luoguP1440

#include<iostream>

#include<cstdio> //scanf...,printf...速度快??

using namespace std;

const int N=2000005;

int q[N],p[N],n,m,x;

int main(){

// ios::sync_with_stdio(0);

// cin.tie(0);

cin>>n>>m;

// cout<<0<<endl;

printf("0\n");

// for(int i=1;i<=n;i++){

// // cin>>a[i];

// scanf("%d",&a[i]);

// }

int head=1,tail=0;

for(int i=1;i<=n-1;i++){

scanf("%d",&x);

while(head<=tail&&q[tail]>x){

tail--;

}

q[++tail]=x;

p[tail]=i;

if(p[head]<=i-m){

head++;

}

// cout<<q[head]<<endl;

printf("%d\n",q[head]); //需使用printf()才不会TLE

}

// cin>>x;

scanf("%d",&x);

return 0;

}

/*

6 2

7 8 1 4 3 2

0

7

7

1

1

3

*/

LC 5402. 绝对差不超过限制的最长连续子数组

class Solution {

public:

int longestSubarray(vector<int>& nums, int limit) {

multiset<int> s; //可重复set

int n=nums.size();

int i=0;

int ans=0;

for(int j=0;j<n;++j){

s.insert(nums[j]);

while(*s.rbegin()-*s.begin()>limit){ //set自动排序，无需求绝对值

s.erase(s.find(nums[i])); //删除队首元素

i++; //头指针后移

}//(因为是任意两元素不满足情况,所以后序无需判断)

ans=max(ans,j-i+1); //头尾指针求队的长度

}

return ans;

}

};

LC1423. 可获得的最大点数

class Solution { //滑动窗口

public:

int maxScore(vector<int>& cardPoints, int k) {

int n=cardPoints.size();

int sum=0,s=0; //sum:n-k个数的和(滑动窗口) s:所有总和

for(int i=0;i<n;++i){

s+=cardPoints[i];

if(i<n-k)

sum+=cardPoints[i]; //sum:尾部留k个空间

}

int ans=sum; //ans=sum

for(int i=0,j=n-k;j<n;++i,++j){ //j<n

sum=sum-cardPoints[i]+cardPoints[j];

ans=min(ans,sum); //滑动窗口和最小(剩余值最大,逆向思维:找最大子数组和)

}

return s-ans; //最后再减:可能取所有元素导致n-k=0 => sum=0

}

};

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