hdu 3974 Assign the task （线段树+树的遍历）

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

Sample Output

Case #1:
-1 
1 
2

代码如下：

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cmath>
  5 #include <algorithm>
  6 #include <queue>
  7 #include <vector>
  8 using namespace std;
  9 const int MAXN = 50500;
 10 int n,Q,t,topboss,cnt;
 11 int head[MAXN],tot;
 12 int start[MAXN],endd[MAXN];
 13 bool used[MAXN];
 14 struct Edge
 15 {
 16     int to,next;
 17 }edge[MAXN];
 18 void init()
 19 {
 20     cnt=0;
 21     tot=0;
 22     memset(head,-1,sizeof head);
 23     memset(used,false,sizeof used);
 24 }
 25 void addedge (int u,int v)
 26 {
 27     edge[tot].to=v;
 28     edge[tot].next=head[u];
 29     head[u]=tot++;
 30 }
 31 void dfs(int u)
 32 {
 33     ++cnt;
 34     start[u]=cnt;
 35     for (int i=head[u];i!=-1;i=edge[i].next)
 36     {
 37         dfs(edge[i].to);
 38     }
 39     endd[u]=cnt;
 40 }
 41 struct Node
 42 {
 43     int l,r,val,lazy;
 44 }segTree[MAXN<<2];
 45 void upDate_Same (int r,int v)
 46 {
 47     if (r)
 48     {
 49         segTree[r].val=v;
 50         segTree[r].lazy=1;
 51     }
 52 }
 53 void push_down (int r)
 54 {
 55     if (segTree[r].lazy)
 56     {
 57         upDate_Same(r<<1,segTree[r].val);
 58         upDate_Same(r<<1|1,segTree[r].val);
 59         segTree[r].lazy=0;
 60     }
 61 }
 62 void buildTree (int i,int l,int r)
 63 {
 64     segTree[i].l=l;
 65     segTree[i].r=r;
 66     segTree[i].val=-1;
 67     segTree[i].lazy=0;
 68     if (l==r)
 69     return ;
 70     int mid =(l+r)>>1;
 71     buildTree(i<<1,l,mid);
 72     buildTree(i<<1|1,mid+1,r);
 73 }
 74 void update (int i,int l,int r,int v)
 75 {
 76     if (segTree[i].l==l&&segTree[i].r==r)
 77     {
 78         upDate_Same(i,v);
 79         return ;
 80     }
 81     push_down(i);
 82     int mid =(segTree[i].l+segTree[i].r)/2;
 83     if (r<=mid) update(i<<1,l,r,v);
 84     else if (l>mid) update(i<<1|1,l,r,v);
 85     else
 86     {
 87         update(i<<1,l,mid,v);
 88         update(i<<1|1,mid+1,r,v);
 89     }
 90 }
 91 int query (int i,int u)
 92 {
 93     if (segTree[i].l==u&&segTree[i].r==u)
 94         return segTree[i].val;
 95     push_down(i);
 96     int mid =(segTree[i].l+segTree[i].r)/2;
 97     if (u<=mid)
 98         return query(i<<1,u);
 99     else
100         return query(i<<1|1,u);
101 }
102 int main()
103 {
104     //freopen("de.txt","r",stdin);
105     cin>>t;
106     int casee=0;
107     while (t--){
108         printf("Case #%d:
",++casee);
109         int u,v;
110         init();
111         scanf("%d",&n);
112         for (int i=0;i<n-1;++i){
113             cin>>u>>v;
114             used[u]=true;
115             addedge(v,u);
116         }
117         for (int i=1;i<=n;++i){
118             if (!used[i]){
119                 dfs(i);
120                 break;
121             }
122         }
123         cin>>Q;
124         buildTree(1,1,cnt);
125         char op[10];
126         while (Q--){
127             scanf("%s",op);
128             if (op[0]=='C'){
129                 scanf("%d",&u);
130                 printf("%d
",query(1,start[u]));
131             }
132             else{
133                 scanf("%d%d",&u,&v);
134                 update(1,start[u],endd[u],v);
135             }
136         }
137     }
138     return 0;
139 }

700ms，有人200ms过了，正在研究。http://vjudge.net/contest/source/7108995 http://vjudge.net/contest/source/6308790