Box
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2374 Accepted Submission(s): 718
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
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题目大意：n个点，然后给出n个点分别的父节点，下边m次操作，move a b,把a放到b里边，b为0，直接放地面，query 问祖先
ac代码
Problem : 2475 ( Box ) Judge Status : Accepted
RunId : 14537757 Language : C++ Author : lwj1994
Code 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);
}
}
}
}
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