Algorithm self-study __ monotonic queue

References:

• https://zhuanlan.zhihu.com/p/346354943

Algorithm Introduction

Monotonic queue can be done in O ( n ) O(n)In the time complexity of$O$$($$n$$)$$In\; a\; sequence\; of\; n$ , each length ismmThe maximum value of the interval of$m\; .$

algorithm thinking

Visual understanding: In each cycle, first check whether the "senior" at the head of the team has graduated , then kick out all the seniors who are inferior to the "freshmen" in the queue, and finally let the freshmen join the team.

The specific algorithm idea can be found in reference materials.

Example 1 P2216 [HAOI2007] ideal square

topic description

There is $a\times A\; matrix\; composed\; of\; integers\; of\; b$ , now please find a$n\times A\; square\; region\; of\; n$ such that the difference between the maximum and minimum values ​​of all numbers in the region is minimized.

input format

First line $3$ integers, respectively representing$a,b,The\; value\; of\; n$ .

The second row to the first $a+1$ line for each line$b$ non-negative integers, representing the number at the corresponding position in the matrix. Each row is separated by a space between two adjacent numbers.

output format

Only one integer, for $a\times$All n × nn \times nin the$b\; matrix$$n\times$The minimum value of the difference between the largest integer and the smallest integer in an$n\; square\; area".$

Example #1

Sample Input #1

5 4 2
1 2 5 6
0 17 16 0
16 17 2 1
2 10 2 1
1 2 2 2

Sample output #1

1

hint

All numbers in the matrix do not exceed $1,000,000,000$。

$100\%$ confidence2$2\le a,b\le 1000,n\le a,n\le b,n\le 100$。

train of thought

General idea: use a monotone queue to find each $n\times The\; maximum\; and\; minimum\; values\; of\; n$ squares, and then traverse each square to update the answer.

Take finding the maximum value within a square as an example. First scan each row of the input matrix, and find the length of each row is nnThe maximum value of the interval of$n$mmax[][] , stored in the matrix . Then scanmmax[][]each legal column in and find each lengthnnThe maximum value of the interval of$n$mmmax[][] , stored in the matrix .

the code

#include<bits/stdc++.h>
using namespace std;

const int maxn = 1e3+5;

int a, b, n;
int mtx[maxn][maxn];
int mmax[maxn][maxn];
int mmmax[maxn][maxn];
int mmin[maxn][maxn];
int mmmin[maxn][maxn];
int ans = 0x7fffffff; 

int main(){
    
    
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	cin>>a>>b>>n;
	for(int i=1;i<=a;i++){
    
    
		for(int j=1;j<=b;j++){
    
    
			cin>>mtx[i][j];
		}
	}
	// 求正方形内最大值
	for(int i=1;i<=a;i++){
    
    
		deque<int> q;
		for(int j=1;j<=b;j++){
    
    
			if(!q.empty() && j-q.front()+1>n) q.pop_front();
			while(!q.empty() && mtx[i][j]>mtx[i][q.back()]) q.pop_back();
			q.push_back(j);
			if(j>=n){
    
    
				mmax[i][j] = mtx[i][q.front()];
			}
		}
	} 
	for(int j=n;j<=b;j++){
    
    
		deque<int> q;
		for(int i=1;i<=a;i++){
    
    
			if(!q.empty() && i-q.front()+1>n) q.pop_front();
			while(!q.empty() && mmax[i][j]>mmax[q.back()][j]) q.pop_back();
			q.push_back(i);
			if(i>=n){
    
    
				mmmax[i][j] = mmax[q.front()][j];
			}
		}
	}
	// 求正方形内最小值 
	for(int i=1;i<=a;i++){
    
    
		deque<int> q;
		for(int j=1;j<=b;j++){
    
    
			if(!q.empty() && j-q.front()+1>n) q.pop_front();
			while(!q.empty() && mtx[i][j]<mtx[i][q.back()]) q.pop_back();
			q.push_back(j);
			if(j>=n){
    
    
				mmin[i][j] = mtx[i][q.front()];
			}
		}
	} 
	for(int j=n;j<=b;j++){
    
    
		deque<int> q;
		for(int i=1;i<=a;i++){
    
    
			if(!q.empty() && i-q.front()+1>n) q.pop_front();
			while(!q.empty() && mmin[i][j]<mmin[q.back()][j]) q.pop_back();
			q.push_back(i);
			if(i>=n){
    
    
				mmmin[i][j] = mmin[q.front()][j];
			}
		}
	}
	// 求最终答案 
	for(int i=n;i<=a;i++){
    
    
		for(int j=n;j<=b;j++){
    
    
			ans = min(ans, mmmax[i][j]-mmmin[i][j]);
		}
	}
	cout<<ans; 
	return 0;
}

Example 2 P2034 select number

topic description

Given a row of $n$ non-negative integers$a_1\cdots a_n$. Now you can choose some of them, but there cannot be more than $k$ consecutive numbers are selected. Your task is to maximize the sum of the selected numbers.

input format

The first line of two integers $n$，$k$。

following $n$ lines, one integer per line represents$a_i$。

output format

Output a value representing the answer.

Example #1

Sample Input #1

5 2
1
2
3
4
5

Sample output #1

12

hint

For $100\%$ of the data,$1\le n\le 100000，1$ ≤ k$1\le k\le n$，$0\le$ number size$\le 1,000,000,000$。

train of thought

This question is a classic example of dynamic programming for monotonic queue optimization .

Convert the topic to: delete several numbers from the topic, and the difference between the serial numbers of any two deleted numbers is less than or equal to k. Under the premise of meeting the requirements of the topic, the sum of the deleted numbers should be as small as possible.

Defining the state dp[i]means: Considering only the first inumber and deleting the first inumber, the smallest sum to be deleted. The state transition equation is:
$$dp[i]=\{\begin{array}{l}a[i],i\le k+1\\ a[i]+\underset{i-k-1\le j\le i-1}{\min }dp[j],i>k+1\end{array}$$

It can be seen that the state transition needs to use the maximum value of the fixed-length interval , so monotonic queue optimization can be used.

the code

#include<bits/stdc++.h>
#define int long long
using namespace std;

const int maxn = 1e5+5; 

int n, k;
int a[maxn];
int dp[maxn];
int sum = 0;
int ans = 0x7fffffff;
deque<int> q;

signed main(){
    
    
	cin>>n>>k;
	k += 1;
	for(int i=1;i<=n;i++){
    
    
		cin>>a[i];
		sum += a[i];
	}
	for(int i=1;i<=n;i++){
    
    
		if(!q.empty() && i>k) dp[i] += dp[q.front()];
		dp[i] += a[i];
		if(!q.empty() && i-q.front()+1>k) q.pop_front();
		while(!q.empty() && dp[i]<dp[q.back()]) q.pop_back();
		q.push_back(i);
	}
	ans = sum;
	for(int i=n-k+1;i<=n;i++){
    
    
		ans = min(ans, dp[i]);
	}
	cout<<sum-ans;
	return 0;
}

Example 3 P2698 [USACO12MAR] Flowerpot S

topic description

The boss needs your help watering the flowers. gives NNCoordinates of$N$$y$ represents the height of the droplet,$x$ means it falls to$The\; position\; of\; the\; x-$ axis.

1 per second per drop$Falling\; at\; a\; speed\; of\; 1$ unit length. You need to place the flower pot at$A\; certain\; position\; on\; the\; x-$ axis makes the 1st$1$ drop of water starts, to the last$1$ drop ends with a time difference of at least$D$。

We believe that as long as the water drop reaches $On\; the\; x-$ axis, it is considered to be caught if it is aligned with the edge of the flowerpot. gives$Coordinates\; of\; N$ drops and$For\; the\; size\; of\; D$ , please calculate the width WWof the smallest flower pot$W$。

input format

first line $2$ integers$N$ and$D$。

Next $N$ lines each line$2$ integers, representing the coordinates of the water drop$(x,y)$。

output format

Only one row $1$ integer, indicating the width of the smallest flowerpot. If it is not possible to construct a pot wide enough that$D$ units of time to catch the water droplets that meet the requirements, then output$-1$。

Example #1

Sample Input #1

4 5
6 3
2 4
4 10
12 15

Sample output #1

2

hint

$100\%$ of data:$1\le N\le 10^5$，$1\le D\le 10^6$，$0\le x,y\le 10^6$ 。

train of thought

Binary answer + monotonic queue

It should be noted that the condition for the head of the team to leave the team is: $The\; difference\; in\; x$ coordinatesis greater than the pot width.

the code

#include<bits/stdc++.h>
using namespace std;

const int maxn = 1e5+5;

struct NODE{
    
    
	int x, y;
	bool operator<(const NODE& a)const{
    
    
		if(x==a.x) return y<a.y;
		else return x<a.x;
	}
};

NODE node[maxn];
int n, d;
int m = 0x7fffffff, M = -1;
int L = 0x7fffffff, R = -1;
int ans = 0x7fffffff;

bool check(int len){
    
    
	deque<int> q1, q2;
	for(int i=1;i<=n;i++){
    
    
		if(!q1.empty() && node[i].x-node[q1.front()].x>len) q1.pop_front();
		if(!q2.empty() && node[i].x-node[q2.front()].x>len) q2.pop_front();
		while(!q1.empty() && node[i].y>node[q1.back()].y) q1.pop_back();
		while(!q2.empty() && node[i].y<node[q2.back()].y) q2.pop_back();
		q1.push_back(i);
		q2.push_back(i);
		if(node[q1.front()].y-node[q2.front()].y >= d) return true;
	}
	return false;
}

int main(){
    
    
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	cin>>n>>d;
	for(int i=1;i<=n;i++){
    
    
		cin>>node[i].x>>node[i].y;
		m = min(m, node[i].y);
		M = max(M, node[i].y);
		L = min(L, node[i].x);
		R = max(R, node[i].x);
	} 
	sort(node+1, node+1+n);
	R = R-L, L = 0;
	if(M-m<d){
    
    
		cout<<-1;
		return 0;
	}
	while(L <= R){
    
    
		int mid = (L+R)>>1;
		if(check(mid)){
    
    
			ans = min(ans, mid);
			R = mid-1;
		}
		else{
    
    
			L = mid+1;
		}
	}
	cout<<ans;
	return 0;
}