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.
InputInput 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.
OutputFor 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
题意：给定一些盒子，以及盒子的嵌套关系，现在有一些操作，可以把盒子移到另外的盒子里。一些询问，问包含x盒子的最外面的盒子是哪个。
思路：题意转化一下，就是森林，Cut(x,y)操作是可以把x为根的子树砍下来，接在y节点下面。 询问Query(x)就是查询x的根。
（模型很裸，还是写一下装进模板里。
像这样维护森林，求根，并查集肯定不够。我们用括号序列来做。
- 那么点x的子树就的区间就是[x,x+N];
- 点y在x的子树里，当且仅当pos[x]<=pos[y]<=pos[y+N]<=pos[x+N]；（pos是点在splay里面的排名。）
- 把x加到y的子树里，即[y,y+N]里面加序列[x,x+N]；可以把后者加到y的后面。
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=100010;
int ch[maxn][2],fa[maxn],Laxt[maxn],Next[maxn],To[maxn],cnt,dd,N;
int get(int x){ return ch[fa[x]][1]==x; }
void rotate(int x)
{
int old=fa[x],fold=fa[old],opt=get(x);
fa[x]=fold; fa[old]=x; fa[ch[x][opt^1]]=old;
ch[old][opt]=ch[x][opt^1];ch[x][opt^1]=old;
if(fold) ch[fold][ch[fold][1]==old]=x;
}
void splay(int x,int y)
{
for(int f;(f=fa[x])!=y;rotate(x)){
if(fa[f]!=y) rotate(get(x)==get(f)?f:x);
}
}
void add(int u,int v)
{
Next[++cnt]=Laxt[u]; Laxt[u]=cnt; To[cnt]=v;
}
void dfs(int u) //按照括号序建spaly，起初全是单链。
{
fa[u]=dd; ch[dd][1]=u; dd=u;
for(int i=Laxt[u];i;i=Next[i]) dfs(To[i]);
fa[u+N]=dd; ch[dd][1]=u+N; dd=u+N;
}
void build()
{
for(int i=Laxt[0];i;i=Next[i]){
dd=0; dfs(To[i]);
}
}
int query(int x)
{
splay(x,0);
int now=x;
while(ch[now][0]) now=ch[now][0];
return now;
}
void move(int a,int b)
{
if(a==b) return ; //不合法1
splay(a,0);
splay(a+N,a);
for(int t=b;t;t=fa[t]) if(ch[a+N][0]==t) return ; //不合法2
int x=ch[a][0],y=ch[a+N][1]; fa[x]=fa[y]=ch[a][0]=ch[a+N][1]=0;
int t=y; while(ch[t][0]) t=ch[t][0];
splay(t,0); fa[x]=t; ch[t][0]=x;//[a,a+N]区间分离出来，合并剩余部分
//右边的最小值放根，再合并，保证平衡。
if(b==0) return ;
splay(b,0);
t=ch[b][1]; while(ch[t][0]) t=ch[t][0];
splay(t,b);// t此时没有左儿子，把[a,a+N]挤进去。
fa[a]=t; ch[t][0]=a;
}
int main()
{
int M,x,y,T=0; char opt[6];
while(~scanf("%d",&N)){
if(T) printf("\n");
else T++;
rep(i,0,N) Laxt[i]=0; cnt=0;
rep(i,0,N+N) ch[i][0]=ch[i][1]=0;
rep(i,1,N){
scanf("%d",&x);
add(x,i);
}
build();
scanf("%d",&M);
rep(i,1,M){
scanf("%s",opt);
if(opt[0]==‘Q‘){
scanf("%d",&x);
printf("%d\n",query(x));
}
else {
scanf("%d%d",&x,&y);
move(x,y);
}
}
}
return 0;
}
原文地址：https://www.cnblogs.com/hua-dong/p/10116963.html