[动态树 LCT] BZOJ 2157 旅游
来源:互联网 发布:nginx cgi 编辑:程序博客网 时间:2024/05/16 07:28
动态树裸题
把边建成一个点就好了
取反,打个标记
#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>using namespace std;inline char nc(){static char buf[100000],*p1=buf,*p2=buf;if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }return *p1++;}inline void read(int &x){char c=nc(),b=1;for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}inline void read(char *s){char c=nc(); int len=0;for (;!(c>='a' && c<='z' || c>='A' && c<='Z');c=nc());for (;c>='a' && c<='z' || c>='A' && c<='Z';s[++len]=c,c=nc()); s[++len]=0; }int n;struct Splay{struct node{int size,idx,rev,fbs;int val,maxv,minv;int sum,minimum,maximum;node *p,*ch[2],*fat;inline void setc(node *c,int d) { ch[d]=c; c->p=this; }inline bool dir() { return p->ch[1]==this; }inline void change(){fbs^=1;swap(minimum,maximum); minimum=-minimum; maximum=-maximum; sum=-sum;if (idx>n)val=-val,minv=-minv,maxv=-maxv;}inline void update() {size=ch[0]->size+ch[1]->size+1;sum=ch[0]->sum+ch[1]->sum+val; minimum=min(minv,min(ch[0]->minimum,ch[1]->minimum));maximum=max(maxv,max(ch[0]->maximum,ch[1]->maximum));}inline void reverse() { rev^=1; swap(ch[0],ch[1]); }inline void pushdown(node *null){if (rev){if (ch[0]!=null) ch[0]->reverse();if (ch[1]!=null) ch[1]->reverse();rev=0;}if (fbs){if (ch[0]!=null) ch[0]->change();if (ch[1]!=null) ch[1]->change();fbs=0;}}}*null,Mem[60005];Splay() { null=Mem; null->p=null->ch[0]=null->ch[1]=null->fat=null; null->sum=0; null->size=0; null->maximum=-1<<30; null->minimum=1<<30; }inline void rot(node *x){if (x==null || x->p==null) return ;bool d=x->dir(); node *p=x->p;if (p->p!=null) p->p->setc(x,p->dir()); else x->p=null;p->setc(x->ch[d^1],d); x->setc(p,d^1); p->update(); x->update(); swap(x->fat,p->fat);}inline void splay(node *x){static node *sta[60005];int pnt=0; node *y=x;while (y!=null) sta[++pnt]=y,y=y->p;for (int i=pnt;i;i--) sta[i]->pushdown(null);while (x->p!=null)if (x->p->p==null)rot(x);elsex->dir()==x->p->dir()?(rot(x->p),rot(x)):(rot(x),rot(x));}inline node *Access(node *x){node *y=null;while (x!=null){splay(x);x->ch[1]->p=null; x->ch[1]->fat=x;x->setc(y,1); y->fat=null;x->update();y=x; x=x->fat;}return y;}inline void Link(node *x,node *y){if (Jud(x,y)) return;Access(x)->reverse();splay(x);x->fat=y;Access(x);}inline void Cut(node *x){Access(x);splay(x);x->ch[0]->p=null; x->ch[0]=null;x->fat=null; x->update();}inline void Cut(node *x,node *y){Access(x)->reverse();Cut(y);}inline node *Root(node *x){Access(x);splay(x);node *y=x;while (y->ch[0]!=null) y=y->ch[0];return y;}inline bool Jud(node *x,node *y){return Root(x)==Root(y);}inline void Change(node *x,node *y){Access(x)->reverse();Access(y)->change();}inline void Change(node *x,int r){Access(x);splay(x);x->val=x->maxv=x->minv=r; x->update();}inline int Sum(node *x,node *y){Access(x)->reverse();return Access(y)->sum;}inline int Max(node *x,node *y){Access(x)->reverse();return Access(y)->maximum;}inline int Min(node *x,node *y){Access(x)->reverse();return Access(y)->minimum;}}LCT;Splay::node *pos[60005];inline void Init(int n){for (int i=1;i<2*n;i++){pos[i]=LCT.Mem+i; pos[i]->p=pos[i]->ch[0]=pos[i]->ch[1]=pos[i]->fat=LCT.null; pos[i]->val=0; pos[i]->maxv=-1<<30; pos[i]->minv=1<<30;pos[i]->maximum=-1<<30; pos[i]->minimum=1<<30; pos[i]->sum=0;pos[i]->idx=i; pos[i]->size=1;}}int main(){int u,v,w,Q;char order[15]={0};freopen("t.in","r",stdin);freopen("t.out","w",stdout);read(n); Init(n);for (int i=1;i<n;i++) {read(u); read(v); read(w);pos[n+i]->val=pos[n+i]->maxv=pos[n+i]->minv=w;LCT.Link(pos[i+n],pos[++u]); LCT.Link(pos[i+n],pos[++v]);}read(Q);while (Q--){read(order); read(u); read(v);if (!strcmp(order+1,"MAX")){printf("%d\n",LCT.Max(pos[++u],pos[++v]));}else if (!strcmp(order+1,"MIN")){printf("%d\n",LCT.Min(pos[++u],pos[++v]));}else if (!strcmp(order+1,"SUM")){printf("%d\n",LCT.Sum(pos[++u],pos[++v]));}else if (!strcmp(order+1,"C")){LCT.Change(pos[u+n],v);}else if (!strcmp(order+1,"N")){LCT.Change(pos[++u],pos[++v]);}}return 0;}
0 0
- [动态树 LCT] BZOJ 2157 旅游
- bzoj 3282 Tree 动态树LCT
- bzoj 2002 Bounce 弹飞绵羊(LCT动态树)
- BZOJ 2049 Cave 洞穴勘测(LCT动态树)
- BZOJ 2157 旅游【裸链剖+线段树
- bzoj 2157: 旅游 树链剖分+线段树
- BZOJ[2157]旅游 树链剖分+线段树
- BZOJ 2157 旅游 树链剖分
- BZOJ 2157 旅游 树链剖分
- BZOJ 2157: 旅游
- [BZOJ]2157: 旅游 树链剖分
- BZOJ 2157: 旅游 树链剖分
- bzoj 2157 旅游
- BZOJ 2157: 旅游 树链剖分
- BZOJ 2157: 旅游
- bzoj 2759 一个动态树好题 LCT 数学
- 动态树~LCT总结
- 动态树LCT总结
- Java连接Redis
- 输入adb shell 时 提示error: more than one device and emulator
- ssm框架搭建三---mybatis打印sql
- AndroidStudio 集成海康威视 Android SDK,即集成萤石Android SDK。
- Swift函数柯里化(Currying)简谈
- [动态树 LCT] BZOJ 2157 旅游
- JBOSS安装配置并搭建本地开发环境
- 求最长回文子串,O(n)复杂度
- Android编程学习笔记 之 SQLite数据存储
- AVL树
- Linux下virtualbox 升级后1908错误,sudo: /etc/init.d/vboxdrv:找不到命令
- 会做题的兔兔——重大4月月赛
- C++ vector::resize
- c语言中宏定义和函数区别