可以直接求闵可夫斯基和,这里介绍一种官解。
按照题面的两个评分尺度构建坐标系,将所有可能的 \(1\) 号节点分值放在平面上,那么最后的答案一定在凸包上。如果我们知道答案的最终方向,那么问题就比较简单了:就是直接寻找特定方向上最远的点(那么就是找一个点使其与选定点的点积最大),只需要让和这个方向点积的贡献最短就行了。虽然我们并不知道方向是什么,不过可以随便钦定一个方向,求点积最大和最小的点,这样一定可以找到凸包上的两个点,这个可以直接树形 DP 做。再以这两个点的连线的垂线为方向,如果凸包上还有点那么一定可以找到新的凸包上的点。反复重复这个过程,知道找出的点积最大和最小的点重合,这时候找齐了凸包上的所有点。找点的时候顺便直接统计答案即可。
这样复杂度是凸包上的点数乘 \(n\) 的,因为做一遍树形 DP 是 \(\mathcal O(n)\),一次只能找到凸包上两个点。不过这样就留下了一个问题:凸包上到底有多少个点,复杂度是否正确?注意到给定的点的坐标都是整数,整点凸包上的点的数量级是我们已经解决的问题。于是总复杂度为 \(\mathcal O(T^{\frac{2}{3}} n)\),其中 \(T \leq 10^7\)。
#include<bits/stdc++.h>
#define int long long
#define ld long double
#define ui unsigned int
#define ull unsigned long long
#define eb emplace_back
#define pb pop_back
#define ins insert
#define mp make_pair
#define pii pair<int,int>
#define fi first
#define se second
#define power(x) ((x)*(x))
#define gcd(x,y) (__gcd((x),(y)))
#define lcm(x,y) ((x)*(y)/gcd((x),(y)))
#define lg(x,y) (__lg((x),(y)))
using namespace std;
namespace FastIO
{
template<typename T=int> inline T read()
{
T s=0,w=1; char c=getchar();
while(!isdigit(c)) {if(c=='-') w=-1; c=getchar();}
while(isdigit(c)) s=(s*10)+(c^48),c=getchar();
return s*w;
}
template<typename T> inline void read(T &s)
{
s=0; int w=1; char c=getchar();
while(!isdigit(c)) {if(c=='-') w=-1; c=getchar();}
while(isdigit(c)) s=(s*10)+(c^48),c=getchar();
s=s*w;
}
template<typename T,typename... Args> inline void read(T &x,Args &...args)
{
read(x),read(args...);
}
template<typename T> inline void write(T x,char ch)
{
if(x<0) x=-x,putchar('-');
static char stk[25]; int top=0;
do {stk[top++]=x%10+'0',x/=10;} while(x);
while(top) putchar(stk[--top]);
putchar(ch);
return;
}
}
using namespace FastIO;
namespace MTool
{
#define TA template<typename T,typename... Args>
#define TT template<typename T>
static const int Mod=998244353;
TT inline void Swp(T &a,T &b) {T t=a;a=b;b=t;}
TT inline void cmax(T &a,T b) {a=a>b?a:b;}
TT inline void cmin(T &a,T b) {a=a<b?a:b;}
TT inline void Madd(T &a,T b) {a=a+b>Mod?a+b-Mod:a+b;}
TT inline void Mdel(T &a,T b) {a=a-b<0?a-b+Mod:a-b;}
TT inline void Mmul(T &a,T b) {a=a*b%Mod;}
TT inline void Mmod(T &a) {a=(a%Mod+Mod)%Mod;}
TT inline T Cadd(T a,T b) {return a+b>=Mod?a+b-Mod:a+b;}
TT inline T Cdel(T a,T b) {return a-b<0?a-b+Mod:a-b;}
TT inline T Cmul(T a,T b) {return a*b%Mod;}
TT inline T Cmod(T a) {return (a%Mod+Mod)%Mod;}
TA inline void Madd(T &a,T b,Args... args) {Madd(a,Cadd(b,args...));}
TA inline void Mdel(T &a,T b,Args... args) {Mdel(a,Cadd(b,args...));}
TA inline void Mmul(T &a,T b,Args... args) {Mmul(a,Cmul(b,args...));}
TA inline T Cadd(T a,T b,Args... args) {return Cadd(Cadd(a,b),args...);}
TA inline T Cdel(T a,T b,Args... args) {return Cdel(Cdel(a,b),args...);}
TA inline T Cmul(T a,T b,Args... args) {return Cmul(Cmul(a,b),args...);}
TT inline T qpow(T a,T b) {int res=1; while(b) {if(b&1) Mmul(res,a); Mmul(a,a); b>>=1;} return res;}
TT inline T qmul(T a,T b) {int res=0; while(b) {if(b&1) Madd(res,a); Madd(a,a); b>>=1;} return res;}
TT inline T spow(T a,T b) {int res=1; while(b) {if(b&1) res=qmul(res,a); a=qmul(a,a); b>>=1;} return res;}
TT inline void exgcd(T A,T B,T &X,T &Y) {if(!B) return X=1,Y=0,void(); exgcd(B,A%B,Y,X),Y-=X*(A/B);}
TT inline T Ginv(T x) {T A=0,B=0; exgcd(x,Mod,A,B); return Cmod(A);}
#undef TT
#undef TA
}
using namespace MTool;
inline void file()
{
freopen(".in","r",stdin);
freopen(".out","w",stdout);
return;
}
bool Mbe;
namespace LgxTpre
{
static const int MAX=10010;
static const int inf=2147483647;
static const int INF=4557430888798830399;
static const int mod=1e9+7;
static const int bas=131;
namespace Geometry
{
struct Point
{
int x,y;
Point(int X=0,int Y=0):x(X),y(Y) {}
inline friend Point operator + (Point a,Point b) {return Point(a.x+b.x,a.y+b.y);}
inline friend Point operator - (Point a,Point b) {return Point(a.x-b.x,a.y-b.y);}
template<typename T> inline friend Point operator * (Point a,T b) {return Point(a.x*b,a.y*b);}
template<typename T> inline friend Point operator / (Point a,T b) {return Point(a.x/b,a.y/b);}
inline friend Point operator * (Point a,Point b) {return Point(a.x*b.x-a.y*b.y,a.y*b.x+a.x*b.y);}
inline Point operator += (Point &T) {*this=*this+T; return *this;}
inline Point operator -= (Point &T) {*this=*this-T; return *this;}
template<typename T> inline Point operator *= (T &x) {*this=*this*x; return *this;}
template<typename T> inline Point operator /= (T &x) {*this=*this/x; return *this;}
inline Point operator *= (Point &T) {*this=*this*T; return *this;}
inline friend bool operator == (Point a,Point b) {return a.x==b.x&&a.y==b.y;}
};
inline int Dot(Point a,Point b) {return a.x*b.x+a.y*b.y;}
inline int Cross(Point a,Point b) {return a.x*b.y-a.y*b.x;}
inline int Norm(Point a) {return sqrt(Dot(a,a));}
}
using namespace Geometry;
int n,k,ans;
Point p[MAX];
vector<int> G[MAX];
inline void lmy_forever()
{
auto Hull=[&](Point P)->pair<Point,Point>
{
auto comp=[&](Point A,Point B)->bool
{
return Dot(P,A)<Dot(P,B);
};
auto dfs=[&](auto dfs,int now)->pair<Point,Point>
{
if(G[now].empty()) return mp(p[now],p[now]);
auto [sl,sr]=dfs(dfs,G[now][0]);
Point vl,vr; vl=vr=sl+sr;
for(int i=1;i<((int)G[now].size());++i)
{
auto [l,r]=dfs(dfs,G[now][i]);
sl+=l,sr+=r;
if(comp(l+r,vl)) vl=l+r;
if(comp(vr,l+r)) vr=l+r;
}
return mp(vl-sr,vr-sl);
};
auto [l,r]=dfs(dfs,1);
cmax(ans,Dot(l,l)),cmax(ans,Dot(r,r));
return mp(l,r);
};
auto solve=[&](auto solve,Point a,Point b)->void
{
if(a==b) return;
int x=(b-a).x,y=(b-a).y;
Point c={-y,x};
auto [_,d]=Hull(c);
if(Dot(d,c)>max(Dot(a,c),Dot(b,c))) solve(solve,a,d),solve(solve,d,b);
};
read(n);
for(int i=1;i<=n;++i)
{
read(k);
if(!k) read(p[i].x,p[i].y);
else for(int j=1;j<=k;++j) G[i].eb(read());
}
auto [a,b]=Hull((Point){1,0});
solve(solve,a,b);
write(ans,'\n');
return;
}
}
bool Med;
int T;
signed main()
{
// file();
fprintf(stderr,"%.3lf MB\n",abs(&Med-&Mbe)/1048576.0);
int Tbe=clock();
LgxTpre::lmy_forever();
int Ted=clock();
cerr<<1e3*(Ted-Tbe)/CLOCKS_PER_SEC<<" ms\n";
return (0-0);
}