# Codeforces 547b Mike and feet (monotonous stack)

Source: Internet
Author: User

Question:
There is a sequence with a length of N, which has a continuous subsequence with a length of 1... n. The smallest value in a continuous subsequence is called the strength of the subsequence of this subsequence,
The maximum strength of a continuous subsequence of each length is required.
Solution:
You can use the stack to obtain the L [I] of each vertex, indicating that the value is smaller than the current position and closest to this point on the left.
Similarly, we can find the R [I], which means the value is smaller than the current position and closest to this point on the right side.
The process of obtaining L [I] is as follows:
Stack S // initially empty
For I = 1 to n
While S is not empty and a [S. Top ()]> = A [I]
Do S. Pop ()
If S is empty
Then l [I] = 0
Otherwise
L [I] = S. Top ()
S. Push (I)
Next, we traverse each position I, so that Len = R [I]-l [I]-1, then ans [Len] = min (A [I], ans [Len])
It can be understood that in R [I]-1 ~ L [I] + 1 in the Len interval, the smallest is a [I], so ans [Len] And a [I] are compared.
Finally, from Len = n-1 ~ 1 traversal, so ans [Len] = max (ANS [Len], a [Len + 1]).
Why?
This is not the case. For example, for example, 1 2 3 2 1, we can only find ans , ANS , ANS , for example, ANS  and ANS  are empty.
This means that if ans [Len] is empty, the value of ANS [Len + 1] is given to it.
Why is it the value of ANS [Len + 1]? First, ANS [Len]> = ans [Len + 1]. Suppose we have a length of Len + 1 and only X ~ In the range of Y, min = 2, then a [Len + 1] = 2,
Assume that a [Len] is greater than a [Len + 1], then there must be x + 1 ~ Y or X ~ The range of the Y-1 and min> 2, but if ans [len-1] is empty, it means there is no such range
Therefore, ANS [Len] = ans [Len + 1] is used.

Code

` 1 #include<bits/stdc++.h> 2 #define lc(a) (a<<1) 3 #define rc(a) (a<<1|1) 4 #define MID(a,b) ((a+b)>>1) 5 #define fin(name)  freopen(name,"r",stdin) 6 #define fout(name) freopen(name,"w",stdout) 7 #define clr(arr,val) memset(arr,val,sizeof(arr)) 8 #define _for(i,start,end) for(int i=start;i<=end;i++) 9 #define FAST_IO ios::sync_with_stdio(false);cin.tie(0);10 using namespace std;11 typedef long long LL;12 const int N=2e5+5;13 const int INF=0x3f3f3f3f;14 const double eps=1e-10;15 16 int l[N],r[N],a[N],ans[N];17 stack<int>sk;18 19 int main(){20     FAST_IO;21     int n;22     cin>>n;23     for(int i=1;i<=n;i++){24         cin>>a[i];25     }26     for(int i=1;i<=n;i++){27         while(!sk.empty()&&a[sk.top()]>=a[i]) 28             sk.pop();29         if(!sk.empty())30             l[i]=sk.top();31         else32             l[i]=0;33         sk.push(i);34     }35     while(!sk.empty()) sk.pop();36     for(int i=n;i>=1;i--){37         while(!sk.empty()&&a[sk.top()]>=a[i])38             sk.pop();39         if(!sk.empty())40             r[i]=sk.top();41         else42             r[i]=n+1;43         sk.push(i);44     }45     for(int i=1;i<=n;i++){46         int len=r[i]-l[i]-1;47         ans[len]=max(a[i],ans[len]);48     }49     for(int i=n-1;i>=1;i--){50         ans[i]=max(ans[i],ans[i+1]);51     }52     for(int i=1;i<=n;i++){53         cout<<ans[i]<<" ";54     }55     cout<<endl;56     return 0;57 }`

Codeforces 547b Mike and feet (monotonous stack)

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