HDOJ 题目2475 Box（link cut tree去点找祖先）

Box

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2374    Accepted Submission(s): 718

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

 

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题目大意：n个点，然后给出n个点分别的父节点，下边m次操作，move a b,把a放到b里边。b为0。直接放地面，query 问祖先

ac代码

Problem : 2475 ( Box )     Judge Status : Accepted
RunId : 14537757    Language : C++    Author : lwj1994
Code Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta
#include<stdio.h>
#include<string.h>
struct LCT
{
    int bef[50050],pre[50050],next[50050][2];
    void init()
    {
        memset(pre,0,sizeof(pre));
        memset(next,0,sizeof(next));
    }
    void rotate(int x,int kind)
    {
        int y,z;
        y=pre[x];
        z=pre[y];
        next[y][!kind]=next[x][kind];
        pre[next[x][kind]]=y;
        next[z][next[z][1]==y]=x;
        pre[x]=z;
        next[x][kind]=y;
        pre[y]=x;
    }
    void splay(int x)
    {
        int rt;
        for(rt=x;pre[rt];rt=pre[rt]);
        if(x!=rt)
        {
            bef[x]=bef[rt];
            bef[rt]=0;
            while(pre[x])
            {
                if(next[pre[x]][0]==x)
                {
                    rotate(x,1);
                }
                else
                    rotate(x,0);
            }
        }
    }
    void access(int x)
    {
        int fa;
        for(fa=0;x;x=bef[x])
        {
            splay(x);
            pre[next[x][1]]=0;
            bef[next[x][1]]=x;
            next[x][1]=fa;
            pre[fa]=x;
            bef[fa]=0;
            fa=x;
        }
    }
    int query(int x)
    {
        access(x);
        splay(x);
        while(next[x][0])
            x=next[x][0];
        return x;
    }
    void cut(int x)
    {
        access(x);
        splay(x);
        bef[next[x][0]]=bef[x];
        bef[x]=0;
        pre[next[x][0]]=0;
        next[x][0]=0;
    }
    void join(int x,int y)
    {
        if(y==0)
            cut(x);
        else
        {
            int tmp;
            access(y);
            splay(y);
            for(tmp=x;pre[tmp];tmp=pre[tmp]);
            if(tmp!=y)
            {
                cut(x);
                bef[x]=y;
            }
        }
    }
}lct;
int main()
{
    int n,flag=0;
    while(scanf("%d",&n)!=EOF)
    {
        int i;
        if(flag)
            printf("\n");
        else
            flag=1;
        for(i=1;i<=n;i++)
        {
            int x;
            scanf("%d",&x);
            lct.bef[i]=x;
        }
        int q;
        lct.init();
        scanf("%d",&q);
        while(q--)
        {
            char str[10];
            scanf("%s",str);
            if(str[0]=='Q')
            {
                int x;
                scanf("%d",&x);
                printf("%d\n",lct.query(x));
            }
            else
            {
                int x,y;
                scanf("%d%d",&x,&y);
                lct.join(x,y);
            }
        }
    }
}