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.

1

5

4 3

3 2

1 3

5 2

5

C 3

T 2 1

 C 3

T 3 2

C 3

Case #1:

-1

1

2

读题之后发现建树是个问题，然后就学习了dfs建树。有一个结论，如果v是u的祖先，那么dfs序st[v]<st[u]&&ed[v]>ed[u]

于是就可以建树了。然后就是打标记查标记

 #include <iostream>

 #include <stdio.h>

 #include <math.h>

 #include <string.h>

 #include <stdlib.h>

 #include <string>

 #include <vector>

 #include <set>

 #include <map>

 #include <queue>

 #include <algorithm>

 #include <sstream>

 #include <stack>

 using namespace std;

 #define FO freopen("in.txt","r",stdin);

 #define rep(i,a,n) for (int i=a;i<n;i++)

 #define per(i,a,n) for (int i=n-1;i>=a;i--)

 #define pb push_back

 #define mp make_pair

 #define all(x) (x).begin(),(x).end()

 #define fi first

 #define se second

 #define SZ(x) ((int)(x).size())

 #define debug(x) cout << "&&" << x << "&&" << endl;

 #define lowbit(x) (x&-x)

 #define mem(a,b) memset(a, b, sizeof(a));

 typedef vector<int> VI;

 typedef long long ll;

 typedef pair<int,int> PII;

 const ll mod=;

 const int inf = 0x3f3f3f3f;

 ll powmod(ll a,ll b) {ll res=;a%=mod;for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}

 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}

 //head

 const int maxn=;

 int _,lazy[maxn<<],st[maxn],ed[maxn],cur,m,vis[maxn],n;

 vector<int> boss[maxn];

 void dfs(int rt) {//建树

     st[rt]=++cur;

     for(int i=;i<boss[rt].size();i++) {

         dfs(boss[rt][i]);

     }

     ed[rt]=cur;

 }

 void pushdown(int rt) {

     if(lazy[rt]!=-) {

         lazy[rt<<]=lazy[rt];

         lazy[rt<<|]=lazy[rt];

         lazy[rt]=-;

     }

 }

 void build(int rt,int L,int R) {

     lazy[rt]=-;

     if(L==R) return;

     int mid=(L+R)>>;

     build(rt<<,L,mid);

     build(rt<<|,mid+,R);

 }

 void updata(int rt,int L,int R,int l,int r,int zhi) {

     if(L>=l&&R<=r) {

         lazy[rt]=zhi;

         return;

     }

     pushdown(rt);

     int mid=(L+R)>>;

     if(l<=mid) updata(rt<<,L,mid,l,r,zhi);

     if(r>mid) updata(rt<<|,mid+,R,l,r,zhi);

 }

 int query(int rt,int L,int R,int pos) {

     if(L==R) return lazy[rt];//单点查

     pushdown(rt);

     int mid=(L+R)>>;

     if(pos<=mid) query(rt<<,L,mid,pos);

     else query(rt<<|,mid+,R,pos);

 }

 int curr=;

 int main() {

     for(scanf("%d",&_);_;_--) {

         printf("Case #%d:\n",curr++);

         cur=;

         mem(boss,);

         mem(vis,);

         scanf("%d",&n);

         int u,v;

         rep(i,,n) {//存关系

             scanf("%d%d",&u,&v);

             boss[v].push_back(u);

             vis[u]=;

         }

         rep(i,,n+) {//找到根

             if(!vis[i]) {

                 dfs(i);

                 break;

             }

         }

         build(,,cur);//建树

         scanf("%d",&m);

         char s[];

         int pos,zhi;

         while(m--) {

             scanf("%s",s);

             if(s[]=='T') {

                 scanf("%d%d",&pos,&zhi);

                 updata(,,cur,st[pos],ed[pos],zhi);//区间st[pos]-ed[pos]是pos的员工

             } else {

                 scanf("%d",&pos);

                 printf("%d\n",query(,,cur,st[pos]));//查pos的任务(ed[pos]就不是了)

             }

         }

     }

 }