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.、
题意：
某公司员工上下级关系呈现树形结构，现在有任务需要分配，某个员工获得任务后，他的所有下属都会转为做该任务。现在需要分配任务以及查询某个员工正在做的任务。
将树形结构用dfs序的方法转变为线性结构，因此他的所有子树内部节点的编号均会在他的进入以及离开编号之间。这样就可以进行线段树区间修改和单点查询了。
1 #include<stdio.h>
2 #include<string.h>
3 const int maxm=5e4+5;
4
5 int head[maxm],nxt[maxm],point[maxm],size;
6 bool f[maxm];
7 int t,stx[maxm],edx[maxm];
8 int st[maxm<<2],ch[maxm<<2];
9
10 void add(int a,int b){
11 point[size]=a;
12 nxt[size]=head[b];
13 head[b]=size++;
14 }
15
16 void dfs(int s){
17 stx[s]=++t;
18 for(int i=head[s];~i;i=nxt[i]){
19 int j=point[i];
20 dfs(j);
21 }
22 edx[s]=t;
23 }
24
25 void pushdown(int o){
26 if(ch[o]!=-1){
27 ch[o<<1]=ch[o];
28 ch[o<<1|1]=ch[o];
29 st[o<<1]=ch[o];
30 st[o<<1|1]=ch[o];
31 ch[o]=-1;
32 }
33 }
34
35 void pushup(int o){
36 if(st[o<<1]==st[o<<1|1])st[o]=st[o<<1];
37 else st[o]=-2;
38 }
39
40 void update(int o,int l,int r,int ql,int qr,int c){
41 if(ql<=l&&qr>=r){
42 ch[o]=c;
43 st[o]=c;
44 return;
45 }
46 pushdown(o);
47 int m=l+((r-l)>>1);
48 if(ql<=m)update(o<<1,l,m,ql,qr,c);
49 if(qr>=m+1)update(o<<1|1,m+1,r,ql,qr,c);
50 pushup(o);
51 }
52
53 int query(int o,int l,int r,int ind){
54 if(st[o]!=-2)return st[o];
55 if(l==r)return st[o];
56 pushdown(o);
57 int m=l+((r-l)>>1);
58 if(ind<=m)return query(o<<1,l,m,ind);
59 return query(o<<1|1,m+1,r,ind);
60 }
61
62 char s[10];
63
64 int main(){
65 int T,cnt=0;
66 scanf("%d",&T);
67 while(T--){
68 memset(head,-1,sizeof(head));
69 size=0;
70 memset(f,0,sizeof(f));
71 t=0;
72 int n;
73 scanf("%d",&n);
74 for(int i=1;i<n;++i){
75 int a,b;
76 scanf("%d%d",&a,&b);
77 f[a]=1;
78 add(a,b);
79 }
80 for(int i=1;i<=n;++i){
81 if(!f[i]){
82 dfs(i);
83 break;
84 }
85 }
86 memset(st,-1,sizeof(st));
87 memset(ch,-1,sizeof(ch));
88 printf("Case #%d:
",++cnt);
89 int m;
90 scanf("%d",&m);
91 for(int i=1;i<=m;++i){
92 scanf("%s",s);
93 if(s[0]=='C'){
94 int a;
95 scanf("%d",&a);
96 printf("%d
",query(1,1,t,stx[a]));
97 }
98 else if(s[0]=='T'){
99 int a,b;
100 scanf("%d%d",&a,&b);
101 update(1,1,t,stx[a],edx[a],b);
102 }
103 }
104 }
105 return 0;
106 }