Hdoj topic 2475 Box (link cut tree go to point to find ancestors)

BoxTime limit:10000/5000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 2374 Accepted Submission (s): 718


Problem Descriptionthere is N boxes on the ground, which is labeled by numbers from 1 to N. The boxes is 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 a operation is illegal, which is, if Box y are contained (directly or indirectly) by box X, or if Y is possible 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 is on the ground.

The picture below shows, 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. The last picture, the root box of box 5 is box 1, and box 3 's root box is itself.
Inputinput contains several test cases.
For each test case, the first line had 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 was on the ground, it's not contained by any box; Otherwise, Box I is directly inside box AI. It is guaranteed, the input state was always correct (No loop exists).
Next line have 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.

Outputfor each query, the output of the result is on a single line. Use a blank line to separate each test case.
Sample Input

15QUERY 1QUERY 2MOVE 2 0MOVE 1 2QUERY 6 4 6 1 04MOVE 4 1QUERY 3MOVE 1 4QUERY 1

Sample Output

11211

Source2008 Asia Regional Chengdu
Recommendlcy | We have carefully selected several similar problems for you:2478 2480 2481 2479 2476 The main idea: N points, and then gives the n points respectively of the parent node, the bottom m operation, move A B, put a in B inside, B is 0, directly on the ground, query asked ancestors AC code

problem:2475 (Box) Judge status:acceptedrunid:14537757 language:c++ author:lwj1994code 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); }        }    }}

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

Hdoj topic 2475 Box (link cut tree go to point to find ancestors)