Codefores 1025D Recovering BST

Posted 2020-12-31 xfanx

给定一个序列，问这个序列是否能够构成一个二叉搜索树，使得任一边连接的点的gcd大于1

区间dp

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;

const int maxn = 710;

int A[maxn];
bool G[maxn][maxn];
bool dp[maxn][maxn][2];

int gcd(int x, int y) {
    return y == 0 ? x : gcd(y, x % y);
}

int main() {
    int n; scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
    for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) {
        G[i][j] = (gcd(A[i], A[j]) > 1);
    }
    for (int len = 1; len <= n; len++) {
        for (int l = 1, r = l + len - 1; l + len - 1 <= n; l++, r++) {
            for (int m = l; m <= r; m++) {
                bool flag = true;
                if (l == m) {
                    flag &= true;
                } else {
                    flag &= dp[l][m - 1][0];
                }
                if (m == r) {
                    flag &= true;
                } else {
                    flag &= dp[m + 1][r][1];
                }
                if (flag) {
                    dp[l][r][1] |= G[m][l - 1];
                    dp[l][r][0] |= G[m][r + 1];
                }
            }
        }
    }

    for (int i = 1; i <= n; i++) {
        bool tag = true;
        if (i == 1) {
            tag &= true;
        } else {
            tag &= dp[1][i - 1][0];
        }
        if (i == n) {
            tag &= true;
        } else {
            tag &= dp[i + 1][n][1];
        }
        if (tag) {
            puts("Yes"); return 0;
        }
    }
    puts("No");
    return 0;
}