Link: http://codeforces.com/problemset/problem/547/ B

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 [1], ANS [3], ANS [5], for example, ANS [2] and ANS [4] 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)