HDU 3974 Assign the task（线段树 单点更新+lazy）

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

 

Source

题意:先输入n个人，然后是n个人的关系(下属 上司)，然后分配任务，每一个上司分到一个任务就把任务给他的所有下属做，即她的下属的任务都是这个任务，任务不会平分，

查询一个人该时刻做的任务

思路:先找到树根，然后dfs求出每一个人控制的编号，然后每次查询时查询该人控制的下属范围，然后就是lazy，记录该人的任务是否已经传递下去了

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<set>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8typedef __int64 ll;#define fre(i,a,b)  for(i = a; i <b; i++)#define free(i,b,a) for(i = b; i >= a;i--)#define mem(t, v)   memset ((t) , v, sizeof(t))#define ssf(n)      scanf("%s", n)#define sf(n)       scanf("%d", &n)#define sff(a,b)    scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pf          printf#define bug         pf("Hi\n")using namespace std;#define INF 0x3f3f3f3f#define N 50005int father[N],num;vector<int>g[N];int le[N],ri[N];struct stud{int le,ri;int va;int lazy;}f[N*4];int cha(int x){if(x!=father[x])father[x]=cha(father[x]);return father[x];}void dfs(int x){    int i;le[x]=++num;fre(i,0,g[x].size()){dfs(g[x][i]);}    ri[x]=num;}void pushdown(int pos){if(f[pos].va==-1) return ;if(f[pos].lazy==0) return ;f[L(pos)].va=f[pos].va;f[R(pos)].va=f[pos].va;f[L(pos)].lazy=f[R(pos)].lazy=1;    f[pos].lazy=0;}void build(int pos,int le,int ri){f[pos].le=le;f[pos].ri=ri;f[pos].lazy=0;f[pos].va=-1;if(le==ri) return ;int mid=MID(le,ri);build(L(pos),le,mid);build(R(pos),mid+1,ri);}void update(int pos,int le,int ri,int va){if(f[pos].le==le&&f[pos].ri==ri){f[pos].va=va;f[pos].lazy=1;return ;}    pushdown(pos);    int mid=MID(f[pos].le,f[pos].ri);    if(mid>=ri)update(L(pos),le,ri,va);else if(mid<le)update(R(pos),le,ri,va);else{update(L(pos),le,mid,va);update(R(pos),mid+1,ri,va);}}int query(int pos,int le){if(f[pos].le==le&&f[pos].ri==le)return f[pos].va;pushdown(pos);int mid=MID(f[pos].le,f[pos].ri);if(mid>=le)return query(L(pos),le);    return query(R(pos),le);}int main(){int i,j,t,m,n,ca=0;sf(t);while(t--){sf(n);fre(i,1,n+1){father[i]=i;g[i].clear();}        int u,v;        m=n-1;        while(m--){sff(u,v);father[u]=v;g[v].push_back(u);}       u=cha(1);       num=0;       dfs(u);       build(1,1,num);       sf(m);       pf("Case #%d:\n",++ca);       char op[10];       while(m--)   {     scanf("%s",op);     if(op[0]=='C')  {  sf(u);  pf("%d\n",query(1,le[u]));  }          else  {   sff(u,v);   update(1,le[u],ri[u],v);  }   }}return 0;}