Assign the task
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4809 Accepted Submission(s): 1909
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
2011 Multi-University Training Contest 14 - Host by FZU
题意：
给定点的上下级关系，规定如果给i分配任务a，那么他的所有下属。都停下手上的工作，开始做a。
操作 T x y 分配x任务y，C x询问x的当前任务；
分析：
难点在于区间更新的区间范围上，其实可以这么想，我们先dfs跑一遍树，每一个结点都有进入的值（dfs序）和出去的值，这个就是区间范围，剩下的就是普通的区间覆盖，单点查询
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;
using namespace std;
typedef long long ll;
const int N = 50000 + 10;
vector<int>g[N];
int fa[N];
int id[N];
int r[N];
int cnt;
void dfs(int cur){
id[cur] = ++cnt;
for (int i = 0; i < g[cur].size(); ++i){
int v = g[cur][i];
dfs(v);
}
r[cur] = cnt; ///右端点
}
int setv[N<<2];
void pushdown(int o){
if (setv[o] != -1){
setv[o<<1] = setv[o<<1|1] = setv[o];
setv[o] = -1;
}
}
void update(int L,int R,int c,int l,int r,int o){
if (L <= l && r <= R){
setv[o] = c;
return;
}
int m = l + r >> 1;
pushdown(o);
if (m >= L){
update(L,R,c, l, m, o << 1);
}
if (m < R){
update(L,R,c,m+1,r,o<<1|1);
}
}
int query(int pos,int l,int r,int o){
if (l == r){
return setv[o];
}
pushdown(o);
int m = l + r >> 1;
if (m >= pos){
return query(pos, l, m, o<<1);
}
else {
return query(pos, m+1,r,o<<1|1);
}
}
int main(){
int T;
int cas=0;
scanf("%d",&T);
while(T--)
{
cnt = 0;
int n;
scanf("%d",&n);
memset(setv,-1,sizeof setv);
memset(fa,-1,sizeof fa);
for (int i = 1; i <= n; ++i) g[i].clear();
int root = 1;
for (int i = 1; i < n; ++i){
int u, v;
scanf("%d %d",&u, &v);
g[v].push_back(u);
fa[u] = v;
}
for (int i = 1; i <= n; ++i){
if (fa[i] == -1) {
root = i;
break;
}
}
dfs(root);
int q;
printf("Case #%d:\n", ++cas);
scanf("%d",&q);
char op[3];
int x,y;
while(q--)
{
scanf("%s%d",op,&x);
if (op[0] == 'C'){
printf("%d\n",query(id[x], 1,n,1));
}
else {
scanf("%d",&y);
update(id[x],r[x],y, 1, n, 1);
}
}
}
return 0;
}
