Assign the task

数据据说很水，用并查集写的

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

#include<stdio.h>#include<string.h>#define maxn 50005using namespace std;struct list{int num;int time;}task[maxn];int pre[maxn];int main(){int t,cnt=1;scanf("%d",&t);while(t--){printf("Case #%d:\n",cnt++);int n,a,b;scanf("%d",&n);if(n==0)continue;memset(task,-1,sizeof(task));memset(pre,0,sizeof(pre));for(int i=0;i<=n;i++)pre[i]=i;for(int i=0;i<n-1;i++){scanf("%d%d",&a,&b);pre[a]=b;}int m,num=1;scanf("%d",&m);for(int i=0;i<m;i++){char ch;scanf(" %c ",&ch);if(ch=='T'){scanf("%d%d",&a,&b);task[a].num=b;task[a].time=num++;}else if(ch=='C'){int r,ans=-1,last=0;scanf("%d",&r);while(r!=pre[r]){if(task[r].time>last){ans=task[r].num;last=task[r].time;}r=pre[r];}if(task[r].time>last){ans=task[r].num;last=task[r].time;}printf("%d\n",ans);}}}return 0;} 

线段树实现（反正不是我写的）：

点击打开链接http://www.cnblogs.com/kuangbin/p/3428460.html

思路（可能我理解的有问题）：

区间更新+点查询

首先前向星存边，之后深搜确定范围（可以这么理解，使每个员工都控制一定的连续区间，在更新过程中每给出一个员工编号就相当于是一段区间），之后建树，更新，查询（只要查询该名员工所控制区间任意元素就行）这些与标准操作差别不大