Codeforces 1025 D - Recovering BST

D - Recovering BST

思路：区间dp

dp[l][r][0]表示l到r之间的数字可以构成一个二叉搜索树，并且以r+1为根节点

dp[l][r][0]表示l到r之间的数字可以构成一个二叉搜索树，并且以l-1为根节点

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 777;
int dp[N][N][2];
int a[N];
bool g[N][N];
int main() {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 0; i <= n+1; i++) {
        for (int j = 0; j <= n+1; j++) {
            g[i][j] = __gcd(a[i], a[j]) != 1;
        }
    }
    for (int i = 1; i <= n+1; i++) dp[i][i-1][0] = dp[i][i-1][1] = 1;
    for (int l = 1; l <= n; l++) {
        for (int i = 1; i+l-1 <= n; i++) {
            int j = i+l-1;
            for (int k = i; k <= j; k++) {
                if(dp[i][k-1][0] && dp[k+1][j][1]) {
                    dp[i][j][0] |= g[k][j+1];
                    dp[i][j][1] |= g[k][i-1];
                }
            }
        }
    }
    if(dp[1][n][0] || dp[1][n][1]) printf("Yes\n");
    else printf("No\n");
    return 0;
}