Recovering BST CodeForces - 1025D (区间dp, gcd)

大意: 给定$n$个数, 任意两个$gcd>1$的数间可以连边, 求是否能构造一棵BST.

数据范围比较大, 刚开始写的$O(n^3\omega(1e9))$竟然T了..优化到$O(n^3)$才过.

思路就是先排个序, 记$L[i][j]$表示区间$[i,j]$是否能组成以$i-1$为根的$BST$, $R[i][j]$为区间$[i,j]$能否组成以$j+1$为根的BST. 然后暴力转移即可.

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <queue>
#define pb push_back
#define REP(i,a,n) for(int i=a;i<=n;++i)
using namespace std;

const int N = 750;
int n, a[N], g[N][N];
vector<int> fac[N];
int L[N][N], R[N][N], c[N][N];

vector<int> calc(int x) {
	vector<int> v;
	for (int i=2; i*i<=x; ++i) if (x%i==0) {
		v.pb(i);
		while (x%i==0) x/=i;
	}
	if (x>1) v.pb(x);
	return v;
}

int main() {
	scanf("%d", &n);
	REP(i,1,n) scanf("%d", a+i);
	sort(a+1,a+1+n);
	REP(i,1,n) {
		fac[i] = calc(a[i]);
		a[i] = 1;
		for (int j:fac[i]) a[i] *= j;
	}
	REP(i,1,n) REP(j,i+1,n) {
		for (int t:fac[i]) {
			if (a[j]%t==0) c[i][j]=c[j][i]=1;
		}
	}
	REP(d,1,n) for (int l=1,r=l+d-1;r<=n;++l,++r) {
		if (d==1) {
			L[l][r] = c[l-1][l];
			R[l][r] = c[l+1][l];
			continue;
		}
		if (L[l+1][r]) {
			if (d==n) return puts("Yes"),0;
			if (l!=1&&!L[l][r]) L[l][r]=c[l-1][l];
			if (r!=n&&!R[l][r]) R[l][r]=c[r+1][l];
		}
		if (R[l][r-1]) {
			if (d==n) return puts("Yes"),0;
			if (l!=1&&!L[l][r]) L[l][r]=c[l-1][r];
			if (r!=n&&!R[l][r]) R[l][r]=c[r+1][r];
		}
		REP(k,l+1,r-1) {
			if (R[l][k-1]&&L[k+1][r]) {
				if (d==n) return puts("Yes"),0;
				if (l!=1&&!L[l][r]) L[l][r]=c[l-1][k];
				if (r!=n&&!R[l][r]) R[l][r]=c[r+1][k];
			}
		}
	}
	puts("No");
}