Logu P2251 Quality Detection and logu p2251 Detection

Logu P2251 Quality Detection and logu p2251 Detection
Background

None

Description

To check the quality of A total of N products on the production line, we first place A score A for each product to indicate its quality, then calculate the score of the product with the worst quality among the first M products. Q [m] = min {A1, A2 ,... am}, and Q [M + 1], Q [m + 2] from 2nd to m + 1... finally, calculate Q [N] from N-M + 1 to n. Perform further evaluation based on Q.

Please obtain the Q sequence as soon as possible.

Input/Output Format

Input Format:

 

Enter two rows in total.

The first line contains two numbers N and M, separated by spaces. The meaning is as described above.

The second line contains N numbers, indicating the quality of N products.

 

Output Format:

 

Output N-M + 1 rows in total.

1st to N-M + 1 rows have one number per row, and the number of I rows Q [I + M-1]. The meaning is as described above.

 

Input and Output sample input sample #1: Copy

10 416 5 6 9 5 13 14 20 8 12

Output example #1: Copy

5555588

Description

[Data range]

30% of data, N <= 1000

100% of data, N <= 100000

100% of data, M <= N, A <= 1 000 000

 

Monotonous queue bare question

 

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<algorithm> 6 #include<deque> 7 #define LL long long  8 #define lb(x)    ((x)&(-x)) 9 using namespace std;10 const int MAXN=1000001;11 inline int read()12 {13     char c=getchar();int x=0,f=1;14     while(c<'0'||c>'9')    {if(c=='-')    f=-1;c=getchar();}15     while(c>='0'&&c<='9')    x=x*10+c-48,c=getchar();return x*f;16 }17 struct node18 {19     int pos,val;20     node(){    pos=val=0;    }21     node(int a,int b){    pos=a,val=b;     }22 };23 deque<node>q;24 int a[MAXN];25 int main()26 {27     int n=read(),m=read();28     for(int i=1;i<=n;i++)29         a[i]=read();30     for(int i=1;i<=n;i++)31     {32         while(q.size()>0&&i-m>=q.front().pos)    q.pop_front();33         while(q.size()>0&&a[i]<=q.back().val)        q.pop_back();34         q.push_back(node(i,a[i]));35         if(i>=m)    printf("%d\n",q.front().val);    36     }37     return 0;38 }