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=0,fh=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')fh=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){s=s*10+(ch-'0');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]!=0)
    {
        rev[x]^=1;rev[l]^=1;rev[r]^=1;
        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=0,y,z,i;Stack[++top]=x;
    for(i=x;!isroot(i);i=father[i])Stack[++top]=father[i];
    for(i=top;i>=1;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=0;
    while(x!=0)
    {
        splay(x);
        tree[x].right=last;
        last=x;x=father[x];
    }
}
void makeroot(int x)
{
    access(x);splay(x);rev[x]^=1;
}
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]=0;tree[u].left=0;
}
int findroot(int x)
{
    access(x);splay(x);
    while(tree[x].left!=0)x=tree[x].left;
    return x;
}
int Findroot(int u,int v)
{
    /*access(v);splay(v);*/access(v);splay(u);
    while(tree[u].right!=0)u=tree[u].right;
    if(u==v)return 1;
    return 0;
}
int main()
{
    int n,i,m,k,x,y,a,tt=0;
    char fh[30];
    while(scanf("%d",&n)!=EOF)
    {
        if(tt)puts("");
        tt++;
        for(i=1;i<=n;i++)
        {
            tree[i].left=tree[i].right=father[i]=rev[i]=cc[i]=0;
        }
        for(i=1;i<=n;i++)
        {
            a=read();
            if(a!=0)
            {
                link(i,a);
                //cc[i]=a;
            }
        }
        m=read();
        for(i=1;i<=m;i++)
        {
            scanf("
%s",fh);
            if(fh[0]=='Q')
            {
                k=read();
                printf("%d
",findroot(k));
            }
            else
            {
                x=read();y=read();
                if(y==0)cut(x,y);
                else if(x!=y)
                {
                    if(Findroot(x,y)==1)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("
");
    }
    return 0;
}