Box
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】
QUERY
QUERY
MOVE
MOVE
QUERY MOVE
QUERY
MOVE
QUERY
【Sample Output】
【题意】
动态地维护一些盒子套盒子的操作，询问根。
【分析】
盒子与盒子的关系可以直观地用树的结构来表示，一个结点下的子结点可以表示大盒子里面直接套着的小盒子。
所以本题就是一个裸的Link-Cut Tree模型了。
关于LCT树，还是推荐Yang Zhe的QTREE论文吧。
动态树是用访问操作来划分树链，对于每一条树链，使用Splay来维护，用深度作为splay的左右关系。
看了很多代码，觉得还是写不好，总觉得别人的用起来不顺，最后是在自己原来Splay的基础上改的。
原本的整棵树是个splay，但是在LCT中，整棵树是由很多棵分散的Splay组合起来的，于是在其中的一些点上加上root标记，表示以这一点为根下面可以形成一棵splay树。多个这样的splay组合完成之后就是一棵LCT了。
后面的代码中加入了输入输出挂。。。。。。
/* ***********************************************
MYID : Chen Fan
LANG : G++
PROG : HDU 2475
************************************************ */ #include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm> using namespace std; #define MAXN 50010 int sons[MAXN][];
int father[MAXN],pathfather[MAXN],data[MAXN];
bool root[MAXN];
int spttail=; void rotate(int x,int w) //rotate(node,0/1)
{
int y=father[x]; sons[y][!w]=sons[x][w];
if (sons[x][w]) father[sons[x][w]]=y;
father[x]=father[y];
if (father[y]&&(!root[y])) sons[father[y]][y==sons[father[y]][]]=x;
sons[x][w]=y;
father[y]=x; if (root[y])
{
root[x]=true;
root[y]=false;
}
} void splay(int x) //splay(node)
{
while(!root[x])
{
if (root[father[x]]) rotate(x,x==sons[father[x]][]);
else
{
int t=father[x];
int w=(sons[father[t]][]==t);
if (sons[t][w]==x)
{
rotate(x,!w);
rotate(x,w);
} else
{
rotate(t,w);
rotate(x,w);
}
}
}
} void access(int v)
{
int u=v;
v=;
while(u)
{
splay(u);
root[sons[u][]]=true;
sons[u][]=v;
root[v]=false;
v=u;
u=father[u];
}
} int findroot(int v)
{
access(v);
splay(v);
while (sons[v][]) v=sons[v][];
//splay(v,0);
return v;
} void cut(int v)
{
access(v);
splay(v);
father[sons[v][]]=;
root[sons[v][]]=true;
sons[v][]=;
} void join(int v,int w)
{
if (!w) cut(v);
else
{
access(w);
splay(w);
int temp=v;
while(!root[temp]) temp=father[temp];
if (temp!=w)
{
cut(v);
father[v]=w;
}
}
} int INT()
{
char ch;
int res;
while (ch=getchar(),!isdigit(ch));
for (res = ch - '';ch = getchar(),isdigit(ch);)
res = res * + ch - '';
return res;
} char CHAR()
{
char ch, res;
while (res = getchar(), !isalpha(res));
while (ch = getchar(), isalpha(ch));
return res;
} int main()
{
//freopen("2475.txt","r",stdin); int n;
double flag=false;
while(scanf("%d",&n)!=EOF)
{
if (flag) printf("\n");
flag=true; memset(father,,sizeof(father));
memset(sons,,sizeof(sons));
for (int i=;i<=n;i++)
{
//scanf("%d",&father[i]);
father[i]=INT();
root[i]=true;
} int m;
m=INT();
for (int i=;i<=m;i++)
{
char s=CHAR();
if (s=='M')
{
int x,y;
x=INT();
y=INT();
join(x,y);
} else
{
int q;
q=INT();
printf("%d\n",findroot(q));
}
}
} return ;
}