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=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 710
int n,a[N];
bool flag[N][N],f[N][N][2];
int gcd(int n,int m){return m==0?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=1;i<=n;i++) a[i]=read();
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
        if (gcd(a[i],a[j])>1) flag[i][j]=1;
    for (int i=1;i<=n+1;i++) f[i][i-1][0]=f[i][i-1][1]=1;
    for (int k=1;k<=n;k++)
        for (int i=1;i<=n-k+1;i++)
        {
            int j=i+k-1;
            for (int d=i;d<=j;d++)
            if (f[i][d-1][1]&&f[d+1][j][0])
            {
                if (flag[i-1][d]) f[i][j][0]=1;
                if (flag[j+1][d]) f[i][j][1]=1;
            }
        }
    for (int i=1;i<=n;i++) if (f[1][i-1][1]&&f[i+1][n][0]) {cout<<"Yes";return 0;}
    cout<<"No";
    return 0;
}

原文地址：https://www.cnblogs.com/Gloid/p/9780193.html