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");
}
Recovering BST CodeForces - 1025D (区间dp, gcd)
原文:https://www.cnblogs.com/uid001/p/11129698.html