There is a company that has N employees(numbered from 1 to N),every employee in the company has a immediate boss (except for the leader of whole company).If you are the immediate boss of someone,that person is your subordinate, and all his subordinates are your subordinates as well. If you are nobody’s boss, then you have no subordinates,the employee who has no immediate boss is the leader of whole company.So it means the N employees form a tree.
The company usually assigns some tasks to some employees to finish.When a task is assigned to someone,He/She will assigned it to all his/her subordinates.In other words,the person and all his/her subordinates received a task in the same time. Furthermore,whenever a employee received a task,he/she will stop the current task(if he/she has) and start the new one.
Write a program that will help in figuring out some employee’s current task after the company assign some tasks to some employee.
Input
The first line contains a single positive integer T( T <= 10 ), indicates the number of test cases.
For each test case:
The first line contains an integer N (N ≤ 50,000) , which is the number of the employees.
The following N - 1 lines each contain two integers u and v, which means the employee v is the immediate boss of employee u(1<=u,v<=N).
The next line contains an integer M (M ≤ 50,000).
The following M lines each contain a message which is either
“C x” which means an inquiry for the current task of employee x
or
"T x y"which means the company assign task y to employee x.
(1<=x<=N,0<=y<=10^9)
Output
For each test case, print the test case number (beginning with 1) in the first line and then for every inquiry, output the correspond answer per line.
Sample Input
1
5
4 3
3 2
1 3
5 2
5
C 3
T 2 1
C 3
T 3 2
C 3
Sample Output
Case #1:
-1
1
2
题意：
一家公司是以老板—>员工—员工的员工这种方式存在的，一个老板可以有多个员工，一旦某个人被分配到任务，他就会去做，并且他的员工会停下手上的任务去做。c x代表问这个人正在做啥，如果没有输出-1，T x y代表让x去做y
题解：
dfs建树，递归末尾记录当前位置的最终儿子下标，之后就是普通线段树了
#include<bits/stdc++.h>
using namespace std;
const int maxn=50005;
struct node
{
int l,r;
}son[maxn];
int change[maxn];
int cnt=0;
int head[maxn];
int n,m;
struct edge
{
int to,next;
}e[maxn];
void add(int x,int y)
{
e[cnt].to=y;
e[cnt].next=head[x];
head[x]=cnt++;
}
int num;
void dfs(int fa)
{
change[fa]=num;
son[change[fa]].l=num;
for(int i=head[fa];~i;i=e[i].next)
{
num++;
dfs(e[i].to);
}
son[change[fa]].r=num;
}
int now[maxn<<2],flag[maxn<<2];
void pushdown(int root)
{
if(flag[root]==-1)
return ;
now[root<<1]=flag[root];
now[root<<1|1]=flag[root];
flag[root<<1]=flag[root];
flag[root<<1|1]=flag[root];
flag[root]=-1;
}
void update(int l,int r,int root,int ql,int qr,int val)
{
if(l>=ql&&r<=qr)
{
now[root]=val;
flag[root]=val;
return ;
}
pushdown(root);
int mid=l+r>>1;
if(mid>=ql)
update(l,mid,root<<1,ql,qr,val);
if(mid<qr)
update(mid+1,r,root<<1|1,ql,qr,val);
}
int query(int l,int r,int root,int q)
{
if(l==r)
return now[root];
pushdown(root);
int mid=l+r>>1;
if(mid>=q)
return query(l,mid,root<<1,q);
else
return query(mid+1,r,root<<1|1,q);
}
int main()
{
int t;
scanf("%d",&t);
int cas=0;
while(t--)
{
scanf("%d",&n);
memset(head,-1,sizeof(head));
cnt=0;
int x,y;
int fa[maxn];
memset(fa,0,sizeof(fa));
for(int i=1;i<n;i++)
scanf("%d%d",&x,&y),add(y,x),fa[x]=1;
num=1;
for(int i=1;i<=n;i++)
if(fa[i]==0)
dfs(i);
memset(now,-1,sizeof(now));
memset(flag,-1,sizeof(flag));
scanf("%d",&m);
printf("Case #%d:\n",++cas);
char s[2];
int pos,val;
while(m--)
{
scanf("%s",s);
if(s[0]=='T')
{
scanf("%d%d",&pos,&val);
pos=change[pos];
int l=son[pos].l,r=son[pos].r;
update(1,n,1,l,r,val);
}
else
{
scanf("%d",&pos);
pos=change[pos];
printf("%d\n",query(1,n,1,pos));
}
}
}
return 0;
}