hdu2475 BOX(splay+dfs序)

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

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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题外话：
在考试中遇到这一题的时候，一开始想到是并查集，但是并查集很显然不能保证几次合并后的祖先是最高级的，只能知道是同一个集合的，之后想到要快速查找祖先有想到了倍增，但事实上在合并更新的操作上会不行。
题解：
这题所需要的操作是合并两棵树，并且快速知道这个树的祖先。
在这里采用splay来维护合并和查找的操作，然后用dfs序来确定一棵树的整体，即一棵树的子节点为[a,a+n]，这样才能保证是在以某个节点为祖先的树中操作。
借鉴了以下的题解（大佬的splay写的真好）

http://blog.csdn.net/afafsdg/article/details/8053656

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<queue>#include<algorithm>#include<stack>#define N 200000struct aa{    int p,ne;}da[N];int tr[N][2],fa[N],tou[N],dd,q;int n,next[N],tmp,size;char s[20];void add(int x,int y){da[++dd].p=y;da[dd].ne=tou[x];tou[x]=dd;}void build(int x){    fa[x]=size;    tr[size][1]=x;    size=x;    for (int i=tou[x];i;i=da[i].ne)build(da[i].p);    fa[x+n]=size;    tr[size][1]=x+n;    size=x+n;}inline void rotate(int x,bool f){    int y=fa[x];    int z=fa[y];    tr[y][!f]=tr[x][f];    fa[tr[x][f]]=y;    fa[x]=z;    if (z)tr[z][tr[z][1]==y]=x;    tr[x][f]=y;    fa[y]=x;}void splay(int x,int k){    int y=fa[x];    while (y!=k){        int z=fa[y];        bool f=(tr[y][0]==x);        if (z!=k&& f==(tr[z][0]==y))rotate(y,f);        rotate(x,f);        y=fa[x];    }}void move(int x,int y){    if (x==y)return;    splay(x,0);splay(x+n,x);    for (int i=y;i;i=fa[i])if (tr[x+n][0]==i)return;    int a=tr[x][0];    int b=tr[x+n][1];    tr[x][0]=tr[x+n][1]=fa[a]=fa[b]=0;    if (a&&b){        while (tr[b][0])b=tr[b][0];        tr[b][0]=a;        fa[a]=b;    }    if (y==0)return;    splay(y,0);    int i;    for (i=tr[y][1];tr[i][0];i=tr[i][0]);    splay(i,y);    tr[i][0]=x;    fa[x]=i;}int query(int x){    splay(x,0);    for (;tr[x][0];x=tr[x][0]);    return x;}int main(){    #ifndef ONLINE_JUDGE    freopen("hdu2475.in","r",stdin);    freopen("hdu2475.out","w",stdout);    #endif    bool xi=0;    while (scanf("%d",&n)!=EOF){        memset(tou,0,sizeof(tou));        memset(next,0,sizeof(next));        memset(da,0,sizeof(da));        tmp=0;dd=0;        if (!xi)xi=1;        else printf("\n");        for (int i=1;i<=n;i++){            int x;            scanf("%d",&x);            if (x)add(x,i);            else {                next[tmp]=i;                tmp=i;            }        }        for (int i=next[0];i;i=next[i])size=0,build(i);        scanf("%d",&q);        for (int i=1;i<=q;i++){            scanf("%s",s);            if (s[0]=='Q'){                int x;                scanf("%d",&x);                printf("%d\n",query(x));            }else{                int x,y;                scanf("%d%d",&x,&y);                move(x,y);            }        }        for (int i=1;i<=2*n;i++)tr[i][1]=tr[i][0]=fa[i]=0;        //printf("\n");    }    return 0;}