# P1637 ternary ascending subsequence (DP + discrete weight tree array)

2021-08-10 07:40:12

P1637 Ternary ascending subsequence (DP+ Discretized weight tree array )

Portal

Ideas ： d p + dp+ dp+ Discretized weight tree array .

It is obvious that d p [ i ] [ j ] dp[i][j] dp[i][j] For a length of i i i With a [ j ] a[j] a[j] The number of subsequences at the end .

There's a transfer equation ： d p [ i ] [ j ] = ∑ k < j , a [ k ] < a [ j ] d p [ i − 1 ] [ k ] dp[i][j]=\sum\limits_{k<j,a[k]<a[j]} dp[i-1][k] dp[i][j]=k<j,a[k]<a[j]dp[i1][k]

Obviously, the time complexity of violence ： O ( n 2 m ) O(n^2m) O(n2m)

because a [ i ] ≤ 2 63 a[i]\leq2^{63} a[i]263, but n ≤ 3 e 4 n\leq 3e4 n3e4 Consider discretization a [ i ] a[i] a[i], Then turn to the weight segment tree to store d p [ i − 1 ] [ k ] dp[i-1][k] dp[i1][k].

First initialize the unary ascending subsequence , Then traverse from front to back ,

There's a transfer equation ： d p [ i ] [ j ] + = q u e r y ( a [ j ] − 1 ) dp[i][j]+=query(a[j]-1) dp[i][j]+=query(a[j]1)

Update again u p d a t e ( a [ j ] , d p [ i − 1 ] [ j ] ) update(a[j],dp[i-1][j]) update(a[j],dp[i1][j]).

Time complexity ： O ( n m l o g n ) O(nmlogn) O(nmlogn), m m m It is a sequence of ascending subsequences .

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
const int N=3e4+5,inf=0x3f3f3f3f,mod=1e9+7;
#define mst(a) memset(a,0,sizeof a)
#define lx x<<1
#define rx x<<1|1
#define reg register
#define PII pair<int,int>
#define fi first
#define se second
int n,m;
ll a[N],b[N],tr[N];
ll dp[4][N];
#define lowbit(x) x&(-x)
void update(int x,int k){
while(x<=m){
tr[x]+=k;
x+=lowbit(x);
}
}
ll query(int x){
ll ans=0;
while(x){
ans+=tr[x];
x-=lowbit(x);
}
return ans;
}
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%lld",&a[i]),b[i]=a[i];
dp[1][i]=1;
}
sort(b+1,b+n+1);
m=unique(b+1,b+n+1)-b-1;
for(int i=1;i<=n;i++) a[i]=lower_bound(b+1,b+m+1,a[i])-b;
for(int i=2;i<=3;i++){
mst(tr);
for(int j=1;j<=n;j++)
{
dp[i][j]+=query(a[j]-1);
update(a[j],dp[i-1][j]);
}
}
ll ans=0;
for(int i=1;i<=n;i++) ans+=dp[3][i];
printf("%lld\n",ans);
return 0;
}


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