Codeforces 1025 D - Recovering BST

时间：2025-03-11 08:34:38

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 = ;

int dp[N][N][];

int a[N];

bool g[N][N];

int main() {

    int n;

    scanf("%d", &n);

    for (int i = ; i <= n; i++) scanf("%d", &a[i]);

    for (int i = ; i <= n+; i++) {

        for (int j = ; j <= n+; j++) {

            g[i][j] = __gcd(a[i], a[j]) != ;

        }

    }

    for (int i = ; i <= n+; i++) dp[i][i-][] = dp[i][i-][] = ;

    for (int l = ; l <= n; l++) {

        for (int i = ; i+l- <= n; i++) {

            int j = i+l-;

            for (int k = i; k <= j; k++) {

                if(dp[i][k-][] && dp[k+][j][]) {

                    dp[i][j][] |= g[k][j+];

                    dp[i][j][] |= g[k][i-];

                }

            }

        }

    }

    if(dp[][n][] || dp[][n][]) printf("Yes\n");

    else printf("No\n");

    return ;

}

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