【JZOJ6287】扭动的树

description

---

analysis

• 区间\(DP\)，首先按照键值排个序，这样保证树的中序遍历就为原序列

• 设\(f[0][i][j]\)表示\([i..j]\)区间作为\([unknown..i-1]\)的右儿子的最大和，\(f[1][i][j]\)就是\([i..j]\)区间作为\([j+1..unknown]\)的左儿子

• 预处理\(f\)的初值是很明显的，然后\(O(n^2log)\)预处理出两两数之间的\(\gcd\)

• 对于一段区间\([i..j]\)，枚举中转点\(k\)，表示\([i..k-1],[k+1,j]\)分别作为\(k\)的左右儿子

• \(k=i\)或\(k=j\)特殊转移，\(i<k<j\)可知\([i..j]\)可由\(f[1][i][k-1],f[0][k+1][j]\)转移得到

• 具体转移到\(0\)或\(1\)取决于\(a[k]\)与\(a[i-1],a[j+1]\)是否符合条件（\(\gcd>1\)）

---

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#define MAXN 305
#define INF 1000000007
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)

using namespace std;

ll f[2][MAXN][MAXN];
ll g[MAXN][MAXN];
ll sum[MAXN];
ll n,ans=-INF;

struct node
{
	ll x,y;
}a[MAXN];

inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
inline ll max(ll x,ll y){return x>y?x:y;}
inline bool cmp(node a,node b){return a.x<b.x;}
inline ll get(ll x,ll y){return sum[y]-sum[x-1];}
inline ll gcd(ll x,ll y){return x%y==0?y:gcd(y,x%y);}
int main()
{
	freopen("T2.in","r",stdin);
	//freopen("tree.in","r",stdin);
	//freopen("tree.out","w",stdout);
	n=read();
	fo(i,0,n)fo(j,0,n)f[0][i][j]=f[1][i][j]=-INF;
	fo(i,1,n)a[i].x=read(),a[i].y=read();
	sort(a+1,a+n+1,cmp);
	fo(i,1,n)fo(j,1,n)g[i][j]=gcd(a[i].x,a[j].x);
	fo(i,1,n)
	{
		sum[i]=sum[i-1]+a[i].y;
		if (i!=1 && g[i][i-1]>1)f[0][i][i]=a[i].y;
		if (i!=n && g[i][i+1]>1)f[1][i][i]=a[i].y;
	}
	fo(len,2,n)
	{
		fo(i,1,n-len+1)
		{
			ll j=i+len-1,tmp;
			fo(k,i,j)
			{
				if (k==i)tmp=f[0][i+1][j]+get(i,j);
				else if (k==j)tmp=f[1][i][j-1]+get(i,j);
				else tmp=f[1][i][k-1]+f[0][k+1][j]+get(i,j);
				if (i!=1 && g[k][i-1]>1)f[0][i][j]=max(f[0][i][j],tmp);
				if (j!=n && g[k][j+1]>1)f[1][i][j]=max(f[1][i][j],tmp);
				if (n==len)ans=max(ans,tmp);
			}
		}
	}
	printf("%lld\n",ans<0?-1ll:ans);
	return 0;
}