LCT用来解决树的形态可变的树上操作查询问题
1 #include<stdio.h> 2 #include<algorithm> 3 using namespace std; 4 struct node{int ch[2],fa,siz;bool rev;}; 5 node sp[200005]; 6 int n,op,fa[200005]; 7 bool isro(int x){return x!=sp[sp[x].fa].ch[0]&&x!=sp[sp[x].fa].ch[1];} 8 void pushup(int x){sp[x].siz=sp[sp[x].ch[0]].siz+sp[sp[x].ch[1]].siz+1;}; 9 void pushdown(int); 10 void Pushdown(int); 11 void rot(int); 12 void splay(int); 13 void access(int); 14 void maker(int x){access(x);splay(x);sp[x].rev^=1;} 15 void link(int x,int y){maker(x);sp[x].fa=y;splay(x);} 16 void cut(int x,int y){maker(x);access(y);splay(y);sp[y].ch[0]=sp[x].fa=0;} 17 int main() 18 { 19 scanf("%d",&n); 20 int i,p,x,y; 21 for(i=1;i<=n;i++) 22 { 23 scanf("%d",&p); 24 fa[i]=sp[i].fa = min(i+p,n+1); 25 sp[i].siz = 1; 26 } 27 sp[n+1].siz = 1; 28 scanf("%d",&op); 29 for(i=1;i<=op;i++) 30 { 31 scanf("%d%d",&p,&x); x++; 32 if(p==1) 33 { 34 maker(n+1); 35 access(x);splay(x); 36 printf("%d ",sp[sp[x].ch[0]].siz); 37 } 38 else 39 { 40 scanf("%d",&y); 41 cut(x,fa[x]); 42 link(x,min(n+1,y+x)); 43 fa[x] = min(n+1,y+x); 44 } 45 } 46 return 0; 47 } 48 void rot(int now) 49 { 50 int fa=sp[now].fa;int fafa=sp[fa].fa; bool isr=(now==sp[fa].ch[1]); 51 if(!isro(fa)){sp[fafa].ch[fa==sp[fafa].ch[1]]=now;} 52 sp[now].fa = fafa; 53 sp[fa].ch[isr] = sp[now].ch[!isr]; 54 sp[sp[now].ch[!isr]].fa = fa; 55 sp[now].ch[!isr] = fa;sp[fa].fa = now; 56 pushup(fa); pushup(now); 57 } 58 void pushdown(int now) 59 { 60 if(sp[now].rev) 61 { 62 int lc = sp[now].ch[0],rc=sp[now].ch[1]; 63 if(lc) sp[lc].rev^=1; 64 if(rc) sp[rc].rev^=1; 65 sp[now].ch[0]=rc,sp[now].ch[1]=lc; 66 sp[now].rev = false; 67 } 68 } 69 void Pushdown(int now) 70 { 71 if(!isro(now)) Pushdown(sp[now].fa); 72 pushdown(now); 73 } 74 void splay(int now) 75 { 76 Pushdown(now); 77 while(!isro(now)) 78 { 79 if(!isro(sp[now].fa)) rot(now); 80 rot(now); 81 } 82 } 83 void access(int x) 84 { 85 for(int t=0;x;t=x,x=sp[x].fa) 86 { 87 splay(x); 88 sp[x].ch[1] = t; 89 } 90 }