Codeforces486 E. LIS of Sequence（LIS唯一性判断）

Posted 2021-11-07 live4m

题意：

解法：

lc[i]表示以a[i]为结尾的LIS长度.
rc[i]表示以a[i]为开头的LIS长度.

设序列的LIS为ma,
如果lc[i]+rc[i]-1==ma,那么说明a[i]可以作为某个LIS的数.

那么如何判断LIS是否一定需要a[i]呢?
mp[x]表示可以作为LIS的数中,lc[i]=x的数个数.
预处理mp[],对于可以在LIS的数a[i],如果mp[lc[i]]==1,那么说明a[i]必选.

code：

#include<bits/stdc++.h>
// #define SYNC_OFF
typedef std::vector<int> VE;
typedef std::pair<int,int> PI;
#define int long long
#define ll long long
#define ull unsigned long long
//fast-coding
#define ST(x) x.begin()
#define ED(x) x.end()
#define RST(x) x.rbegin()
#define RED(x) x.end()
#define CL(x) x.clear();
#define all(a,n) a+1,a+1+n
#define ff(i,n) for(ll i=1;i<=n;i++)
#define rff(i,n) for(ll i=n;i>=1;i--)
#define fff(i,n) for(ll i=0;i<n;i++)
#define rfff(i,n) for(ll i=n-1;i>=0;i--)
#define SC(x) scanf("%s",x)
#define SL(x) strlen(x)
#define pss(a) push_back(a)
#define ps(a) push(a)
#define SZ(x) (int)x.size()
#define pee puts("");
#define eee putchar(' ');
#define re readdd()
#define pr(a) printtt(a)
int readdd(){int x=0,f=1;char c=getchar();//
while(!isdigit(c)&&c!='-')c=getchar();
if(c=='-')f=-1,c=getchar();
while(isdigit(c))x=x*10+c-'0',c=getchar();
return f*x;}
void printtt(int x){if(x<0)putchar('-'),x=-x;//
if(x>=10)printtt(x/10);putchar(x%10+'0');}
int gcd(int a,int b){return b==0?a:gcd(b,a%b);}//
int ppow(int a,int b,int mod){a%=mod;//
int ans=1%mod;while(b){if(b&1)ans=(long long)ans*a%mod;
a=(long long)a*a%mod;b>>=1;}return ans;}
bool addd(int a,int b){return a>b;}
int lowbit(int x){return x&-x;}
const int dx[4]={0,0,1,-1};
const int dy[4]={1,-1,0,0};
bool isdigit(char c){return c>='0'&&c<='9';}
bool Isprime(int x){
    for(int i=2;i*i<=x;i++)if(x%i==0)return 0;
    return 1;
}
void ac(int x){if(x)puts("YES");else puts("NO");}
//
using namespace std;
// const int mod=998244353;
const int mod=1e9+7;
const int maxm=2e6+5;
const int BIT_maxm=2e6+5;
int lc[maxm],rc[maxm];
int ans[maxm];
int a[maxm];
int n;
void solve(){
    n=re;
    ff(i,n)a[i]=re;
    VE temp;
    //lc[i]表示以a[i]结尾的最大长度
    //求最长上升子序列
    ff(i,n){
        int p=lower_bound(ST(temp),ED(temp),a[i])-ST(temp);
        if(p==SZ(temp))temp.pss(a[i]);
        else temp[p]=a[i];
        lc[i]=p+1;
    }
    int ma=SZ(temp);
    //rc[i]表示以a[i]开头的最大长度
    //反转后求最长下降子序列
    //取反变为求最长上升子序列
    reverse(all(a,n));
    ff(i,n)a[i]=-a[i];
    CL(temp);
    ff(i,n){
        int p=lower_bound(ST(temp),ED(temp),a[i])-ST(temp);
        if(p==SZ(temp))temp.pss(a[i]);
        else temp[p]=a[i];
        rc[i]=p+1;
    }
    reverse(all(rc,n));//记得翻转回去
    //
    ff(i,n)ans[i]=1;
    map<int,int>mp;
    ff(i,n){
        if(lc[i]+rc[i]-1==ma){
            ans[i]=2;
            mp[lc[i]]++;
        }
    }
    ff(i,n){
        if(lc[i]+rc[i]-1==ma&&mp[lc[i]]==1){
            ans[i]=3;
        }
    }
    //
    ff(i,n){
        pr(ans[i]);
    }
}
void Main(){
    // #define MULTI_CASE
    #ifdef MULTI_CASE
    int T;cin>>T;while(T--)
    #endif
    solve();
}
void Init(){
    #ifdef SYNC_OFF
    ios::sync_with_stdio(0);cin.tie(0);
    #endif
    #ifndef ONLINE_JUDGE
    freopen("../in.txt","r",stdin);
    freopen("../out.txt","w",stdout);
    #endif
}
signed main(){
    Init();
    Main();
    return 0;
}

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