Hdu 2475-Box LCT,动态树

Box

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2813    Accepted Submission(s): 821

Problem Description

There are N boxes on the ground, which are labeled by numbers from 1 to N. The boxes are magical, the size of each one can be enlarged or reduced arbitrarily.
Jack can perform the “MOVE x y” operation to the boxes: take out box x; if y = 0, put it on the ground; Otherwise, put it inside box y. All the boxes inside box x remain the same. It is possible that an operation is illegal, that is, if box y is contained (directly or indirectly) by box x, or if y is equal to x.
In the following picture, box 2 and 4 are directly inside box 6, box 3 is directly inside box 4, box 5 is directly inside box 1, box 1 and 6 are on the ground.

The picture below shows the state after Jack performs “MOVE 4 1”:

Then he performs “MOVE 3 0”, the state becomes:

During a sequence of MOVE operations, Jack wants to know the root box of a specified box. The root box of box x is defined as the most outside box which contains box x. In the last picture, the root box of box 5 is box 1, and box 3’s root box is itself.

 

Input

Input contains several test cases.
For each test case, the first line has an integer N (1 <= N <= 50000), representing the number of boxes.
Next line has N integers: a1, a2, a3, ... , aN (0 <= ai <= N), describing the initial state of the boxes. If ai is 0, box i is on the ground, it is not contained by any box; Otherwise, box i is directly inside box ai. It is guaranteed that the input state is always correct (No loop exists).
Next line has an integer M (1 <= M <= 100000), representing the number of MOVE operations and queries.
On the next M lines, each line contains a MOVE operation or a query:
1.  MOVE x y, 1 <= x <= N, 0 <= y <= N, which is described above. If an operation is illegal, just ignore it.
2.  QUERY x, 1 <= x <= N, output the root box of box x.

 

Output

For each query, output the result on a single line. Use a blank line to separate each test case.

 

Sample Input

2
0 1
5
QUERY 1
QUERY 2
MOVE 2 0
MOVE 1 2
QUERY 1
6
0 6 4 6 1 0
4
MOVE 4 1
QUERY 3
MOVE 1 4
QUERY 1

 

Sample Output

1
1
2

1
1

 

Source

2008 Asia Regional Chengdu

 

Recommend

lcy

 

题解：

LCT关于子树的问题。

判断y是否在x的子树中特别。。。

 #include<bits/stdc++.h>
 using namespace std;
 #define MAXN 50010
 struct node
 {
     int left,right;
 }tree[MAXN];
 int father[MAXN],rev[MAXN],Stack[MAXN],cc[MAXN];
 int read()
 {
     int s=,fh=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')fh=-;ch=getchar();}
     while(ch>=''&&ch<=''){s=s*+(ch-'');ch=getchar();}
     return s*fh;
 }
 int isroot(int x)
 {
     return tree[father[x]].left!=x&&tree[father[x]].right!=x;
 }
 void pushdown(int x)
 {
     int l=tree[x].left,r=tree[x].right;
     if(rev[x]!=)
     {
         rev[x]^=;rev[l]^=;rev[r]^=;
         swap(tree[x].left,tree[x].right);
     }
 }
 void rotate(int x)
 {
     int y=father[x],z=father[y];
     if(!isroot(y))
     {
         if(tree[z].left==y)tree[z].left=x;
         else tree[z].right=x;
     }
     if(tree[y].left==x)
     {
         father[x]=z;father[y]=x;tree[y].left=tree[x].right;tree[x].right=y;father[tree[y].left]=y;
     }
     else
     {
         father[x]=z;father[y]=x;tree[y].right=tree[x].left;tree[x].left=y;father[tree[y].right]=y;
     }
 }
 void splay(int x)
 {
     int top=,y,z,i;Stack[++top]=x;
     for(i=x;!isroot(i);i=father[i])Stack[++top]=father[i];
     for(i=top;i>=;i--)pushdown(Stack[i]);
     while(!isroot(x))
     {
         y=father[x];z=father[y];
         if(!isroot(y))
         {
             if((tree[y].left==x)^(tree[z].left==y))rotate(x);
             else rotate(y);
         }
         rotate(x);
     }
 }
 void access(int x)
 {
     int last=;
     while(x!=)
     {
         splay(x);
         tree[x].right=last;
         last=x;x=father[x];
     }
 }
 void makeroot(int x)
 {
     access(x);splay(x);rev[x]^=;
 }
 void link(int u,int v)
 {
     makeroot(u);/*access(u);*/father[u]=v;splay(u);//////////////
 }
 void cut(int u,int v)
 {
     /*makeroot(u);*//*access(u);*/access(u);splay(u);father[tree[u].left]=;tree[u].left=;
 }
 int findroot(int x)
 {
     access(x);splay(x);
     while(tree[x].left!=)x=tree[x].left;
     return x;
 }
 int Findroot(int u,int v)
 {
     /*access(v);splay(v);*/access(v);splay(u);
     while(tree[u].right!=)u=tree[u].right;
     if(u==v)return ;
     return ;
 }
 int main()
 {
     int n,i,m,k,x,y,a,tt=;
     char fh[];
     while(scanf("%d",&n)!=EOF)
     {
         if(tt)puts("");
         tt++;
         for(i=;i<=n;i++)
         {
             tree[i].left=tree[i].right=father[i]=rev[i]=cc[i]=;
         }
         for(i=;i<=n;i++)
         {
             a=read();
             if(a!=)
             {
                 link(i,a);
                 //cc[i]=a;
             }
         }
         m=read();
         for(i=;i<=m;i++)
         {
             scanf("\n%s",fh);
             if(fh[]=='Q')
             {
                 k=read();
                 printf("%d\n",findroot(k));
             }
             else
             {
                 x=read();y=read();
                 if(y==)cut(x,y);
                 else if(x!=y)
                 {
                     if(Findroot(x,y)==)continue;
                     cut(x,y);link(x,y);
                     /*access(y);splay(x);
                     int t=x;
                     while(tree[t].right!=0)t=tree[t].right;
                     if(t!=y){cut(x,y);link(x,y);}*/
                 }
                 /*if(x!=0&&y!=0)
                 {
                     if(Findroot(y,x)==1)continue;
                 }
                 if(cc[x]!=0)
                 {
                     cut(x,cc[x]);
                     //access(x);access(cc[x]);splay(cc[x]);father[tree[cc[x]].left]=0;tree[cc[x]].left=0;splay(cc[x]);//father[x]=tree[cc[x]].left=0;
                 }
                 cc[x]=y;
                 if(cc[x]!=0)
                 {
                     //link(x,cc[x]);
                     access(x);father[x]=cc[x];//splay(x);
                 }*/
             }
         }
         //printf("\n");
     }
     return ;
 }