[Assign the task][dfs序+线段树]

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Assign the task

Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7144 Accepted Submission(s): 2708

Problem Description

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：给一棵树，有单点查询和区间更新两种操作

题解：通过dfs序转化为区间问题，使用线段树维护即可

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<stack>
  4 using namespace std;
  5 #define debug(x) cout<<"["<<#x<<"]"<<" is "<<x<<endl;
  6 typedef long long ll;
  7 const int maxn=5e4+5;
  8 int n,head[maxn],in[maxn],out[maxn],cnt,tim;
  9 bool f[maxn];
 10 struct node{
 11     int l;
 12     int r;
 13     int val;
 14     int lazy;
 15 }N[maxn<<2];
 16 struct edge{
 17     int q;
 18     int w;
 19     int nex;
 20 }e[maxn<<1];
 21 void pushup(int rt){
 22     N[rt].val=N[rt<<1].val+N[(rt<<1)|1].val;
 23     return;
 24 }
 25 void build(int L,int R,int rt){
 26     N[rt].l=L;
 27     N[rt].r=R;
 28     N[rt].lazy=0;
 29     if(L==R){
 30         N[rt].val=-1;
 31         return;
 32     }
 33     build(L,(L+R)/2,rt<<1);
 34     build((L+R)/2+1,R,(rt<<1)|1);
 35 }
 36 void pushdown(int rt){
 37     if(N[rt].lazy){
 38         N[rt<<1].lazy=N[rt].lazy;
 39         N[(rt<<1)|1].lazy=N[rt].lazy;
 40         N[rt<<1].val=N[(rt<<1)|1].val=N[rt<<1].lazy;
 41         N[rt].lazy=0;
 42     }
 43 }
 44 void update(int L,int R,int rt,int L1,int R1,int c){
 45     if(L1<=L&&R1>=R){
 46         N[rt].val=c;
 47         N[rt].lazy=c;
 48         return;
 49     }
 50     int mid=(L+R)/2;
 51     pushdown(rt);
 52     if(L1<=mid)update(L,mid,rt<<1,L1,R1,c);
 53     if(R1>mid)update(mid+1,R,(rt<<1)|1,L1,R1,c);
 54 }
 55 int query(int L,int R,int rt,int id){
 56     if(L==R){
 57         return N[rt].val;
 58     }
 59     int mid=(L+R)/2;
 60     pushdown(rt);
 61     if(id<=mid)return query(L,mid,rt<<1,id);
 62     else return query(mid+1,R,(rt<<1)|1,id);
 63 }
 64 void adde(int x,int y){
 65     e[cnt].q=y;
 66     e[cnt].w=x;
 67     e[cnt].nex=head[y];
 68     head[y]=cnt++;
 69 }
 70 void dfs(int u){
 71     in[u]=++tim;
 72     for(int i=head[u];i!=-1;i=e[i].nex){
 73         int v=e[i].w;
 74         dfs(v);
 75     }
 76     out[u]=tim;
 77 }
 78 int main(){
 79     int t;
 80     int case1=0;
 81     scanf("%d",&t);
 82     while(t--){
 83         scanf("%d",&n);
 84         tim=0;
 85         cnt=0;
 86         for(int i=1;i<=n;i++){f[i]=0;head[i]=-1;}
 87         build(1,n,1);
 88         for(int i=1;i<n;i++){
 89             int a,b;
 90             scanf("%d%d",&a,&b);
 91             adde(a,b);
 92             f[a]=1;
 93         }
 94         for(int i=1;i<=n;i++){if(!f[i])dfs(i);}
 95         int k;
 96         scanf("%d",&k);
 97         printf("Case #%d:
",++case1);
 98         while(k--){
 99             char ch[10];
100             scanf("%s",ch);
101             if(ch[0]=='T'){
102                 int x,xx;
103                 scanf("%d%d",&x,&xx);
104                 update(1,n,1,in[x],out[x],xx);
105             }
106             else{
107                 int x;
108                 scanf("%d",&x);
109                 printf("%d
",query(1,n,1,in[x]));
110             }
111         }
112     }
113     return 0;
114 }

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