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】
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】
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，虽然坑了这么久，但是真的要我写的话，不知道该从哪开写。还是推荐Yang Zhe的QTREE论文吧。
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。。。。。。
1 /* ***********************************************
2 MYID : Chen Fan
3 LANG : G++
4 PROG : HDU 2475
5 ************************************************ */
6
7 #include <iostream>
8 #include <cstdio>
9 #include <cstring>
10 #include <algorithm>
11
12 using namespace std;
13
14 #define MAXN 50010
15
16 int sons[MAXN][2];
17 int father[MAXN],pathfather[MAXN],data[MAXN];
18 bool root[MAXN];
19 int spttail=0;
20
21 void rotate(int x,int w) //rotate(node,0/1)
22 {
23 int y=father[x];
24
25 sons[y][!w]=sons[x][w];
26 if (sons[x][w]) father[sons[x][w]]=y;
27 father[x]=father[y];
28 if (father[y]&&(!root[y])) sons[father[y]][y==sons[father[y]][1]]=x;
29 sons[x][w]=y;
30 father[y]=x;
31
32 if (root[y])
33 {
34 root[x]=true;
35 root[y]=false;
36 }
37 }
38
39 void splay(int x) //splay(node)
40 {
41 while(!root[x])
42 {
43 if (root[father[x]]) rotate(x,x==sons[father[x]][0]);
44 else
45 {
46 int t=father[x];
47 int w=(sons[father[t]][0]==t);
48 if (sons[t][w]==x)
49 {
50 rotate(x,!w);
51 rotate(x,w);
52 } else
53 {
54 rotate(t,w);
55 rotate(x,w);
56 }
57 }
58 }
59 }
60
61 void access(int v)
62 {
63 int u=v;
64 v=0;
65 while(u)
66 {
67 splay(u);
68 root[sons[u][1]]=true;
69 sons[u][1]=v;
70 root[v]=false;
71 v=u;
72 u=father[u];
73 }
74 }
75
76 int findroot(int v)
77 {
78 access(v);
79 splay(v);
80 while (sons[v][0]) v=sons[v][0];
81 //splay(v,0);
82 return v;
83 }
84
85 void cut(int v)
86 {
87 access(v);
88 splay(v);
89 father[sons[v][0]]=0;
90 root[sons[v][0]]=true;
91 sons[v][0]=0;
92 }
93
94 void join(int v,int w)
95 {
96 if (!w) cut(v);
97 else
98 {
99 access(w);
100 splay(w);
101 int temp=v;
102 while(!root[temp]) temp=father[temp];
103 if (temp!=w)
104 {
105 cut(v);
106 father[v]=w;
107 }
108 }
109 }
110
111 int INT()
112 {
113 char ch;
114 int res;
115 while (ch=getchar(),!isdigit(ch));
116 for (res = ch - ‘0‘;ch = getchar(),isdigit(ch);)
117 res = res * 10 + ch - ‘0‘;
118 return res;
119 }
120
121 char CHAR()
122 {
123 char ch, res;
124 while (res = getchar(), !isalpha(res));
125 while (ch = getchar(), isalpha(ch));
126 return res;
127 }
128
129 int main()
130 {
131 //freopen("2475.txt","r",stdin);
132
133 int n;
134 double flag=false;
135 while(scanf("%d",&n)!=EOF)
136 {
137 if (flag) printf("\n");
138 flag=true;
139
140 memset(father,0,sizeof(father));
141 memset(sons,0,sizeof(sons));
142 for (int i=1;i<=n;i++)
143 {
144 //scanf("%d",&father[i]);
145 father[i]=INT();
146 root[i]=true;
147 }
148
149 int m;
150 m=INT();
151 for (int i=1;i<=m;i++)
152 {
153 char s=CHAR();
154 if (s==‘M‘)
155 {
156 int x,y;
157 x=INT();
158 y=INT();
159 join(x,y);
160 } else
161 {
162 int q;
163 q=INT();
164 printf("%d\n",findroot(q));
165 }
166 }
167 }
168
169 return 0;
170 }