486E-LIS of Sequence (LIS)

486E-LIS of Sequence (LIS)

Question: Give a sequence with a length of n and ask each number about the role of LIS (longest increasing subsequence) in the sequence. There are three types:

1. This number does not appear in any LIS.

2. This number appears in at least one but not all LIS.

3. This number appears in all LIS

Solution: The LIS algorithm of nlgn can obtain the LIS length up [I] ending at each I position. The number of LIS is actually a dag. Just find the number of unique values in a layer. After the LIS algorithm, the system scans forward from the back and maintains the maximum length. In the middle, it can be judged that a length has several values. If some lengths have multiple positions, they all belong to 2, if a length has only one position, it belongs to 3, and the rest are non-LIS elements. When judging multiple numbers, a certain number of num [I] can relax its length position. Make sure that it is smaller than the number of help [up [I + 1] at this time, this ensures that he is in LIS (the last element of LIS is specific ).

Code:

#include 
 
  #include 
  
   #include #include 
   
    #define inf 10000000using namespace std;const int Max=100010;int num[Max];int up[Max];int down[Max];int D[Max];void getLIS(int num[],int n){    int last=0;    for(int i=0; i
    
     last) D[last = up[i]]=num[i];        D[up[i]-1]=num[i];    }}int help[Max];int ne[Max];int ans[Max];int main(){    int n;    while(cin>>n)    {        for(int i=0; i
     
      =0; i--)            ma=max(ma,up[i]);        for(int i=n-1; i>=0; i--)        {            if(up[i]==ma)            {                if(help[up[i]]==0)                    ans[i]=3,ne[ma]=i;                else                {                    ans[i]=2;                    ans[ne[ma]]=2;                    ne[ma]=i;                }                help[up[i]]=num[i];            }            else            {                if(num[i]>=help[up[i]+1])                    ans[i]=1;                else                {                    if(help[up[i]]==0)                        ans[i]=3,ne[up[i]]=i;                    else                    {                        ans[i]=2;                        ans[ne[up[i]]]=2;                        ne[up[i]]=i;                    }                    help[up[i]]=num[i];                }            }        }        for(int i=0; i