HDU 3947 Assign the task
http://acm.hdu.edu.cn/showproblem.php?pid=3974
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.
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)
5
4 3
3 2
1 3
5 2
5
C 3
T 2 1
C 3
T 3 2
C 3
-1
1
2
代码:
#include <bits/stdc++.h>
using namespace std; const int maxn = 1e5 + 10;
int T, N;
int head[maxn];
int cnt, tot;
int st[maxn], en[maxn];
bool used[maxn]; struct Edge {
int to, nx;
}edge[maxn]; struct Node {
int val;
int l, r;
int lazy;
}node[maxn * 4]; void init() {
cnt = 0;
tot = 0;
memset(head, -1, sizeof(head));
} void addedge(int u, int v) {
edge[tot].to = v;
edge[tot].nx = head[u];
head[u] = tot ++;
} void dfs(int u) {
cnt ++;
st[u] = cnt;
for(int i = head[u]; i != -1; i = edge[i].nx)
dfs(edge[i].to);
en[u] = cnt;
} void Update_Same (int r, int v) {
if(r) {
node[r].val = v;
node[r].lazy = 1;
}
} void push_down(int r) {
if(node[r].lazy) {
Update_Same(r * 2, node[r].val);
Update_Same(r * 2 + 1, node[r].val);
node[r].lazy = 0;
}
} void Build(int i, int l, int r) {
node[i].l = l;
node[i].r = r;
node[i].val = -1;
node[i].lazy = 0;
if(l == r) return ;
int mid = (l + r) / 2;
Build(i * 2, l, mid);
Build(i * 2 + 1, mid + 1, r);
} void update(int i, int l, int r, int v) {
if(node[i].l == l && node[i].r == r) {
Update_Same(i, v);
return;
}
push_down(i);
int mid = (node[i].l + node[i].r) / 2;
if(r <= mid) update(i * 2, l, r, v);
else if(l > mid) update(i * 2 + 1, l, r, v);
else {
update(i * 2, l, mid, v);
update(i * 2 + 1, mid + 1, r, v);
}
} int query(int i, int u) {
if(node[i].l == u && node[i].r == u)
return node[i].val;
push_down(i);
int mid = (node[i].l + node[i].r) / 2;
if(u <= mid) return query(i * 2, u);
else return query(i * 2 + 1, u);
} int main() {
scanf("%d", &T);
for(int t = 1; t <= T; t ++) {
printf("Case #%d:\n", t);
int u, v;
memset(used, false, sizeof(used));
init();
scanf("%d", &N);
for(int i = 1; i < N; i ++) {
scanf("%d%d", &u, &v);
used[u] = true;
addedge(v, u);
}
for(int i = 1; i <= N; i ++) {
if(!used[i]) {
dfs(i);
break;
}
}
Build(1, 1, cnt);
char op[10];
int M;
scanf("%d", &M);
while(M --) {
scanf("%s", op);
if(op[0] == 'C') {
scanf("%d", &u);
printf("%d\n", query(1, st[u]));
} else {
scanf("%d%d", &u, &v);
update(1, st[u], en[u], v);
}
}
}
return 0;
}
线段树 模板题 链表存图 区间修改单点查询 今天小目标之一把这个搞清楚可以自己写粗来
何生枷锁
HDU 3947 Assign the task的更多相关文章
- HDU 3974 Assign the task 并查集/图论/线段树
Assign the task Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?p ...
- HDU 3974 Assign the task 暴力/线段树
题目链接: 题目 Assign the task Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/O ...
- HDU 3974 Assign the task(简单线段树)
Assign the task Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) ...
- hdu 3874 Assign the task
Assign the task Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) ...
- HDU 3974 Assign the task
Assign the task Problem Description There is a company that has N employees(numbered from 1 to N),ev ...
- HDU 3974 Assign the task(DFS序+线段树单点查询,区间修改)
描述There is a company that has N employees(numbered from 1 to N),every employee in the company has a ...
- 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 comp ...
- hdu 3974 Assign the task(dfs序上线段树)
Problem Description There is a company that has N employees(numbered from 1 to N),every employee in ...
- HDU 3974 Assign the task(dfs建树+线段树)
题目大意:公司里有一些员工及对应的上级,给出一些员工的关系,分配给某员工任务后,其和其所有下属都会进行这项任务.输入T表示分配新的任务, 输入C表示查询某员工的任务.本题的难度在于建树,一开始百思不得 ...
随机推荐
- MacOS远程Windows提示:远程桌面连接无法验证您希望连接的计算机的身份
解决方法: 1.在Windows端,运行输入 “gpedit.msc”,打开本地组策略编辑器 2.依次打开[计算机配置]→[管理模板]→[windows组件]→[远程桌面服务]→[远程桌面会话主机]→ ...
- 实现Github和Coding仓库等Git服务托管更新
如何使Github.Coding.Gitee 码云 同时发布更新,多个不同Git服务器之间同时管理部署发布提交 缘由 因为在Github上托管的静态页面访问加载速度较为缓慢,故想在Coding上再建一 ...
- php学习----基本介绍及数据类型
php 官方手册:http://php.net/manual/zh/ 1.PHP(全称 Hypertext Preprocessor,超文本预处理器的字母缩写)是一种服务器端脚本语言,它可嵌入到 HT ...
- 2802:小游戏利用bfs来实现
之前使用的是递归的方法来解决的问题,后来有点想用bfs(宽度优先搜索来尝试一下的想法,在网上看到有人使用了dfs(深度优先搜索)更加坚定了自己的这种想法. 这个方法首先是以顶点的四组开始,加入那些没有 ...
- C语言 大小写字母转换
//凯鲁嘎吉 - 博客园 http://www.cnblogs.com/kailugaji/ 方法1: #include<stdio.h> #include<stdlib.h> ...
- CORS(跨域资源共享)详解及与JSONP的对比
上篇讲解的JSONP的跨域请求方式,但是在浏览器的支持及请求方式上有局限性,接下来将介绍一种新的跨域请求方式CORS. CORS是一个W3C标准,全称是"跨域资源共享"(Cross ...
- 【Teradata System】How Teradata uses MPP Systems
内存分配 内存初始化时将分配给操作系统和Vprocs,内存不使用部分的90%做为FSG (File Segment Cache) ,由PDE对FSG进行管理. FSG Cache:缓存常驻内存的dat ...
- javascript:针对yyyy-MM-dd hh:mm:ss格式的毫秒转换.
javascript:针对yyyy-MM-dd hh:mm:ss格式的毫秒转换. 时间 2015-11-07 22:18:39 HideHai's blog 原文 http://www.hideh ...
- D - Counting Test Gym - 101532D 字符串
Yousef has a string s that is used to build a magical string w by repeating the string s infinitely ...
- Linux 与 Unix 到底有什么不同?区别在哪?
Linux 与 Unix 到底有什么不同?区别在哪? 如果你是一名20多岁或30多岁的软件开发人员,那么你已成长在一个由Linux主导的世界中.数十年来,它一直是数据中心的重要参与者,尽管很难找到明确 ...