题解：【CF1843F】 Omsk Metro
都这么强？都会倍增？
首先结论是肯定的：记路径上最大子段和为 \(maxx\)，最小子段和为 \(minx\)，有解当且仅当 \(minx \leq k \leq maxx\)，所以要做的是维护最大/最小前缀和，最大/最小后缀和，区间和，最大/最小子段和。其他几篇题解都证过了我们就直接跳过。看到这种动态加点和路径信息维护不应该直接无脑上 LCT 吗？注意 splay 区间翻转的时候还要交换前缀和与后缀和即可，和 splay 维护区间子段和是一样的，可以参考 GSS6。因为保证路径合法，所以动态加点是假的，先把操作离线建好整棵树，这样 LCT 需要实现的只有 split 操作。于是我们做到了 \(\mathcal O(n \log n)\)，甚至可以强制在线。
#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=200010;
static const int inf=2147483647;
static const int INF=4557430888798830399;
static const int mod=1e9+7;
static const int bas=131;
struct node
{
int sum,minx,premin,sufmin,maxx,premax,sufmax;
node(int Val=0) {sum=minx=premin=sufmin=maxx=premax=sufmax=Val;}
inline friend node operator + (node l,node r)
{
node res;
res.sum=l.sum+r.sum;
res.premin=min(l.premin,l.sum+r.premin),res.premax=max(l.premax,l.sum+r.premax);
res.sufmin=min(r.sufmin,r.sum+l.sufmin),res.sufmax=max(r.sufmax,r.sum+l.sufmax);
res.minx=min({l.minx,r.minx,l.sufmin+r.premin}),res.maxx=max({l.maxx,r.maxx,l.sufmax+r.premax});
return res;
}
};
namespace Link_Cat_Tree
{
int fa[MAX],ch[MAX][2],tag[MAX],siz[MAX],val[MAX];
node lmy[MAX];
inline bool get(int i) {return ch[fa[i]][1]==i;}
inline bool noroot(int i) {return ch[fa[i]][get(i)]==i;}
inline void pushup(int i) {siz[i]=siz[ch[i][0]]+siz[ch[i][1]]+1,lmy[i]=lmy[ch[i][0]]+(node){val[i]}+lmy[ch[i][1]];}
inline void down(int i) {if(!i) return; tag[i]^=1,Swp(ch[i][0],ch[i][1]),Swp(lmy[i].premin,lmy[i].sufmin),Swp(lmy[i].premax,lmy[i].sufmax);}
inline void pushdown(int i) {if(tag[i]) down(ch[i][0]),down(ch[i][1]),tag[i]=0;}
inline void pushall(int i) {if(noroot(i)) pushall(fa[i]); pushdown(i);}
inline void rotate(int x)
{
int y=fa[x],z=fa[y];
bool k1=get(x),k2=get(y);
if(noroot(y)) ch[z][k2]=x; fa[x]=z;
ch[y][k1]=ch[x][!k1],fa[ch[x][!k1]]=y;
ch[x][!k1]=y,fa[y]=x;
pushup(y),pushup(x);
}
inline void splay(int x)
{
pushall(x);
while(noroot(x))
{
int y=fa[x];
if(noroot(y)) (get(x)^get(y))?rotate(x):rotate(y);
rotate(x);
}
}
inline void access(int x) {for(int y=0;x;y=x,x=fa[x]) splay(x),ch[x][1]=y,pushup(x);}
inline void makeroot(int x) {access(x),splay(x),down(x);}
inline int findroot(int x) {access(x),splay(x); while(ch[x][0]) pushdown(x),x=ch[x][0]; return splay(x),x;}
inline void split(int x,int y) {makeroot(x),access(y),splay(y);}
inline void link(int x,int y) {makeroot(x); if(findroot(y)!=x) fa[x]=y;}
inline void cut (int x,int y) {makeroot(x); if(findroot(y)==x&&fa[y]==x&&!ch[y][0]) fa[y]=ch[x][1]=0,pushup(x);}
}
using namespace Link_Cat_Tree;
int T,n,all,con;
char c;
int x,y,z;
struct qry{int x,y,z,ans;}q[MAX];
inline void lmy_forever()
{
read(T);
while(T--)
{
for(int i=1;i<=all;++i) fa[i]=ch[i][0]=ch[i][1]=tag[i]=siz[i]=val[i]=0,lmy[i]=(node){0};
read(n),all=1,con=0;
val[1]=1,siz[1]=1,lmy[1]=(node){1};
for(int i=1;i<=n;++i)
{
do c=getchar(); while(c!='+'&&c!='?');
if(c=='+') read(x,z),++all,val[all]=z,lmy[all]=(node){z},fa[all]=x,siz[all]=1;
if(c=='?') ++con,read(q[con].x,q[con].y,q[con].z);
}
for(int i=1;i<=con;++i) {split(q[i].x,q[i].y); if(lmy[q[i].y].minx<=q[i].z&&q[i].z<=lmy[q[i].y].maxx) puts("YES"); else puts("NO");}
}
return;
}
}
bool Med;
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);
}
