【HDU2019多校】Beauty Of Unimodal Sequence (LIS + 思维）

You are given an array of

n

InputThere are multiple test cases.

Each case starts with a line containing a positive integer $n (n \leq 3 \times 10^{5})$ OutputFor each test case, output two lines, the first of which depicts the lexicographically smallest one in longest beautiful sequences and the second of which depicts the lexicographically largest one in longest beautiful sequences.Sample Input

7
1 4 2 5 7 4 6
3
1 2 3
3
3 2 1

Sample Output

1 2 4 5 6
1 3 4 5 7
1 2 3
1 2 3
1 2 3
1 2 3

SOLUTION:


设两个数组 f1 i 表示从i位向后的最长下降子序列  
f2 i 表示从i位向后的最长单峰子序列
通过这两个函数我们可以从前到后，为了保证字典序最小，那就能取就取。贪心的构造
对于字典序最大，就把原数组倒过来在做一次就行了


CODE:

#include<bits/stdc++.h>
using namespace std;
const int N=3e5+5;
int a[N],b[N],pre[N][2],suf[N][2];
int lowbit(int x){return x&(-x);}
int asc[N],desc[N];
void addasc(int x,int v)
{
    for(int i=x;i<N;i+=lowbit(i))
        asc[i]=max(asc[i],v);
}
int queryasc(int x)
{
    int ans=0;
    for(int i=x;i;i-=lowbit(i))
        ans=max(ans,asc[i]);
    return ans;
}
void adddesc(int x,int v)
{
    for(int i=x;i;i-=lowbit(i))
        desc[i]=max(desc[i],v);
}
int querydesc(int x)
{
    int ans=0;
    for(int i=x;i<N;i+=lowbit(i))
        ans=max(ans,desc[i]);
    return ans;
}
struct node
{
    int id,p;
    bool operator< (const node& aa)const
    {
        if(p!=aa.p)
            return p<aa.p;
        return id<aa.id;
    }
};
vector<node>vec;
int num[N],ans[N][2];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(asc,0,sizeof(asc));
        memset(desc,0,sizeof(desc));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]),b[i]=a[i];
            for(int j=0;j<=1;j++)
                pre[i][j]=suf[i][j]=ans[i][j]=0;
        }
        sort(b+1,b+1+n);
        int all=unique(b+1,b+1+n)-b-1;
        for(int i=1;i<=n;i++)
            a[i]=lower_bound(b+1,b+1+all,a[i])-b;
        for(int i=1;i<=n;i++)
        {
            int k=queryasc(a[i]-1);
            pre[i][0]=k+1;
            k=querydesc(a[i]+1);
            pre[i][1]=max(pre[i][0],k+1);
            addasc(a[i],pre[i][0]),adddesc(a[i],max(pre[i][1],pre[i][0]));
        }
        memset(asc,0,sizeof(asc));
        memset(desc,0,sizeof(desc));
        int mx=0;
        for(int i=n;i>=1;i--)
        {
            int k=queryasc(a[i]-1);
            suf[i][0]=k+1;
            k=querydesc(a[i]+1);
            suf[i][1]=max(suf[i][0],k+1);
            addasc(a[i],suf[i][0]),adddesc(a[i],max(suf[i][1],suf[i][0]));

            num[i]=max(pre[i][0]+suf[i][0]-1,max(pre[i][1]+suf[i][0]-1,pre[i][0]+suf[i][1]-1));
            mx=max(num[i],mx);
        }
        int f=0;
        int cnt=0;
        int now=-1;
        for(int i=1;i<=n;i++)
        {
            if(!f)
            {
                if(a[i]>now && suf[i][1]+cnt==mx)
               {
                    if(suf[i][0]+cnt == mx)f=1;
                    now=a[i];ans[++cnt][0]=i;

               }
            }
            else
            {
               if(a[i]<now && suf[i][0]+cnt==mx)
               {
                   now=a[i];ans[++cnt][0]=i; f=1;
               }
            }



        }
        cnt=f=0;
        for(int i=n;i>=1;i--)
        {
            if(num[i]==mx)
            {
                if(f)
                {
                    if(cnt+pre[i][0]==mx)
                        ans[mx-cnt][1]=i,cnt++;
                }
                else
                {
                    if(cnt+1==suf[i][0])
                        ans[mx-cnt][1]=i,cnt++;
                    else if(pre[i][0]+cnt==mx)
                        ans[mx-cnt][1]=i,cnt++;
                }
            }
        }
        for(int i=1;i<=mx;i++)
            printf("%d%c",ans[i][0],i==mx?'\n':' ');
        for(int i=1;i<=mx;i++)
            printf("%d%c",ans[i][1],i==mx?'\n':' ');
    }
    return 0;
}
/*//////
10
10 4 5 8 4 6 8 5 3 10

*/

【HDU2019多校】Beauty Of Unimodal Sequence (LIS + 思维）

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