Codeforces 1025D（区间dp）

　　容易想到设f[i][j][k]为i~j区间以k为根是否能构成bst。这样是O(n⁴)的。考虑将状态改为f[i][j][0/1]表示i~j区间以i-1/j+1为根能否构成bst。显然如果是i-1作为根的话i~j区间都在它的右子树，所以转移时枚举右子树的根并判断是否合法，j+1类似。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 710
int n,a[N];
bool flag[N][N],f[N][N][];
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("d.in","r",stdin);
    freopen("d.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    for (int i=;i<=n;i++) a[i]=read();
    for (int i=;i<=n;i++)
        for (int j=;j<=n;j++)
        if (gcd(a[i],a[j])>) flag[i][j]=;
    for (int i=;i<=n+;i++) f[i][i-][]=f[i][i-][]=;
    for (int k=;k<=n;k++)
        for (int i=;i<=n-k+;i++)
        {
            int j=i+k-;
            for (int d=i;d<=j;d++)
            if (f[i][d-][]&&f[d+][j][])
            {
                if (flag[i-][d]) f[i][j][]=;
                if (flag[j+][d]) f[i][j][]=;
            }
        }
    for (int i=;i<=n;i++) if (f[][i-][]&&f[i+][n][]) {cout<<"Yes";return ;}
    cout<<"No";
    return ;
}