思路：

我们设f[i][j][0/1]表示i~j这一段节点的根节点在左边或右边的最大权值和。
只要枚举根节点k，像区间DP那样去转移就行了

$code$

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

#define ll long long

using namespace std;

const ll MAXN = 310;

ll n, g[MAXN][MAXN], s[MAXN], f[MAXN][MAXN][2];

struct node {
    
	ll key, val;
}a[MAXN];

ll gcd(ll x, ll y) {
    
	if(y == 0) return x;
	else return gcd(y, x % y);
}

bool cmp(node x, node y) {
    
	return x.key < y.key;
}

int main() {
    
	freopen("tree.in", "r", stdin);
	freopen("tree.out", "w", stdout);
	scanf("%lld", &n);
	for(ll i = 1; i <= n; i ++)
		scanf("%lld%lld", &a[i].key, &a[i].val);
	memset(f, -0x3f, sizeof(f));
	sort(a + 1, a + 1 + n, cmp);
	for(ll i = 1; i <= n; i ++)
		for(ll j = 1; j <= n; j ++) g[i][j] = gcd(a[i].key, a[j].key);
	for(ll i = 1; i <= n; i ++) s[i] = a[i].val + s[i - 1];
	for(ll i = 1; i <= n; i ++) {
    
		if(i != 1 && g[i][i - 1] != 1) f[i][i][0] = a[i].val;
		if(i != n && g[i][i + 1] != 1) f[i][i][1] = a[i].val;
	}
	ll ans = - 1e18;
	for(ll l = 2; l <= n; l ++) {
    
		for(ll i = 1; i + l - 1 <= n; i ++) {
    
			ll j = i + l - 1;
			for(ll k = i; k <= j; k ++) {
    
				ll sum;
				if(k == i) sum = f[i + 1][j][0] + s[j] - s[i - 1];
				else if(k == j) sum = f[i][j - 1][1] + s[j] - s[i - 1];
				else sum = f[i][k - 1][1] + f[k + 1][j][0] + s[j] - s[i - 1];
				if(i != 1 && g[k][i - 1] != 1) f[i][j][0] = max(f[i][j][0], sum);
				if(j != n && g[k][j + 1] != 1) f[i][j][1] = max(f[i][j][1], sum);
				if(l == n) ans = max(ans, sum);
			}
		}
	}
	if(ans == - 1e18) printf("-1");
	else printf("%lld", ans);
	return 0;
}