描述
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
有N个雇员，然后有N-1行每行u，v代表v是u的老板，当老板有任务时，他会让他的下属一起做这个任务
有M个操作，操作C代表询问雇员X当前的任务是什么，没有输出-1，操作T代表给雇员X一个任务Y
题解
首先可以看出一个关系树，我们知道dfs时间戳可以知道每个节点为根开始访问的第一个节点（本身）s和最后一个节点e
然后我们用线段树去维护
对于操作C，我们直接单点s[x]查询
对于操作T，我们通过dfs序，区间[s[x],e[x]]修改延迟标记
代码
1 #include<stdio.h>
2 #include<string.h>
3 #include<vector>
4 using namespace std;
5
6 const int N=5e4+5;
7
8 int n,ans;
9 int s[N],e[N],vis[N],tot;
10 int lazy[N<<2];
11 vector<int>G[N];
12
13 void PushDown(int rt)
14 {
15 if(lazy[rt]!=-1)
16 {
17 lazy[rt<<1]=lazy[rt<<1|1]=lazy[rt];
18 lazy[rt]=-1;
19 }
20 }
21
22 void Update(int L,int R,int task,int l,int r,int rt)
23 {
24 if(L<=l&&r<=R)
25 {
26 lazy[rt]=task;
27 return;
28 }
29 int mid=(l+r)>>1;
30 PushDown(rt);
31 if(L<=mid)Update(L,R,task,l,mid,rt<<1);
32 if(R>mid)Update(L,R,task,mid+1,r,rt<<1|1);
33 }
34
35 void Query(int L,int l,int r,int rt)
36 {
37 if(L==l&&L==r)
38 {
39 ans=lazy[rt];
40 return;
41 }
42 int mid=(l+r)>>1;
43 PushDown(rt);
44 if(L<=mid)Query(L,l,mid,rt<<1);
45 else Query(L,mid+1,r,rt<<1|1);
46 }
47
48 void dfs(int u)
49 {
50 tot++;
51 s[u]=tot;
52 for(auto X:G[u])dfs(X);
53 e[u]=tot;
54 }
55
56 int main()
57 {
58 int t,q,u,v,x,y,o=1;
59 char op[3];
60 scanf("%d",&t);
61 while(t--)
62 {
63 printf("Case #%d:\n",o++);
64 memset(vis,0,sizeof vis);
65 memset(lazy,-1,sizeof lazy);
66 scanf("%d",&n);
67 for(int i=1;i<=n;i++)G[i].clear();
68 for(int i=1;i<n;i++)
69 {
70 scanf("%d%d",&u,&v);
71 vis[u]=1;
72 G[v].push_back(u);
73 }
74 ///dfs序
75 for(int i=1;i<=n;i++)
76 if(!vis[i]){tot=0;dfs(i);break;}
77 scanf("%d",&q);
78 for(int i=0;i<q;i++)
79 {
80 scanf("%s",op);
81 if(op[0]==‘C‘)
82 {
83 scanf("%d",&x);
84 Query(s[x],1,n,1);
85 printf("%d\n",ans);
86 }
87 if(op[0]==‘T‘)
88 {
89 scanf("%d%d",&x,&y);
90 Update(s[x],e[x],y,1,n,1);
91 }
92 }
93 }
94 return 0;
95 }
原文地址：https://www.cnblogs.com/taozi1115402474/p/9286724.html