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.
InputThe 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)OutputFor 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
译文：
有一个公司有N个员工（从1到N），公司里的每个员工都有一个直接的老板（除了整个公司的领导），如果你是某人的顶头上司，那个人就是你的下属，他的下属也都是你的下属。如果你不是任何人的老板，那么你就没有下属，没有直接上司的员工是整个公司的领导，这就意味着N员工组成了一棵树。
公司通常会指派一些任务给一些员工完成，当分配给某人的任务时，他/她会把它分配给所有的下属，换句话说，这个人和他/她的下属同时接受一项任务。此外，每当员工收到任务时，他/她将停止当前任务（如果他/她有）并启动新任务。
写，有助于在公司给某个员工分配任务后，计算出员工的当前任务。
题解：这道题不知道为什么归类为线段树这一类里面，感觉没什么关系，就是模拟吧，好像就过了，向上找直到最终祖先为止，
然后找时间戳最迟
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstring> 6 const int MAXN=50007; 7 char s[20]; 8 int num,fa[MAXN],zhi[MAXN],now[MAXN],u,v,t,ans,mm,x,y,n,q,fzy=0; 9 using namespace std; 10 void query(int x) 11 { 12 if (x==-1) return; 13 if (zhi[x]!=-1&&now[x]>mm) 14 { 15 mm=now[x]; 16 ans=zhi[x]; 17 } 18 query(fa[x]); 19 } 20 int main() 21 { 22 scanf("%d",&t); 23 while (t--) 24 { 25 printf("Case #%d: ",++fzy); 26 memset(fa,-1,sizeof(fa)); 27 memset(zhi,-1,sizeof(zhi)); 28 memset(now,-1,sizeof(now)); 29 num=0; 30 scanf("%d",&n); 31 for (int i=1,v,u;i<n;i++) 32 { 33 scanf("%d%d",&v,&u); 34 fa[v]=u; 35 } 36 scanf("%d",&q); 37 for (int i=1;i<=q;i++) 38 { 39 scanf("%s",&s); 40 if (s[0]=='C') 41 { 42 scanf("%d",&x); 43 ans=-1;mm=-10; 44 query(x); 45 printf("%d ",ans); 46 } 47 else 48 { 49 scanf("%d%d",&x,&y); 50 zhi[x]=y; 51 now[x]=++num; 52 } 53 } 54 } 55 }
的。