NBOJv2 1034 Salary Inequity(DFS序+线段树区间更新区间(最值)查询)
Problem 1034: Salary Inequity
Time Limits: 10000 MS Memory Limits: 200000 KB
64-bit interger IO format: %lld Java class name: Main
Description
There is a large company of N employees. With the exception of one employee, everyone has a direct supervisor. The employee with no direct supervisor is indirectly a supervisor of all other employees in the company.
We say that employee X is a subordinate of employee Y if either Y is the direct supervisor of X, or the direct supervisor of X is a subordinate of Y .
One day, the HR department decides that it wants to investigate how much inequity there is in the company with respect to salaries. For a given employee, the inequity of the employee is the difference between the
minimum salary of that employee and all his/her subordinates and the maximum salary of that employee and all his/her subordinates.
HR wants to be able to compute the inequity for any employee quickly. However, this is complicated by the fact that an employee will sometimes give himself/herself, along with all his/her subordinates, a raise.
Can you help?
Input
The first line of your input file contains a number T representing the number of companies you will be analyzing for inequity. T will be at most 20.
For each company, there will be a line containing an integer N, representing the number of employees at the company. Each employee is assigned an ID which is a unique integer from 1 to N. The next line will contain
N − 1 integers. The Kth integer in that line is the ID of the direct supervisor of employee (K + 1). The next line will contain N integers, the Kth integer in this line being the salary of employee K. The next line contains an integer Q, the number of events
that you will need to process. There are two types of events to process - raises and inequity queries. In the event of a raise, the line will start with the letter R followed by the ID of the employee and an integer representing the increase in salary for
that employee and all his/her subordinates. In the event of an inequity query, the line will start with the letter Q followed by the ID of the employee for whom inequity needs to be determined.
2 <= N <= 1,000,000
1 <= Q <= 10,000
For every employee K, the ID of his/her supervisor will be less than K. Initial salaries will
range from 1 to 1,000. No raise will exceed 1,000.
Output
For each inequity query, print the inequity of the employee on its own line.
Sample Input
1
5
1 1 2 2
10 6 8 4 5 7
Q 2
Q 3
R 4 2
Q 2
Q 1
R 2 4
Q 1
Output for Sample Input
2
0
1
5
2
线段树是对于连续的区间操作,但是如果是一棵树,显然是仅符合线段树思想但并不连续,DFS标号构造一个序列,然后再用线段树处理……好题……,输入外挂优化后200MS+,一开始pushdown时左右子树的add忘记加了WA几次…
代码:
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define MM(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=1000010;
int total,in[N],OUT[N];
int arr[N];
int n;
int ref[N];
int head[N],cnt;
struct info
{
int to;
int pre;
}E[N<<1];
struct seg
{
int l,mid,r;
int maxm,minm,add;
};
seg T[N<<2];
void add(int s,int t)
{
E[cnt].to=t;
E[cnt].pre=head[s];
head[s]=cnt++;
}
void dfs(int now,int fa)
{
in[now]=++total;
ref[total]=now;
for (int i=head[now]; ~i; i=E[i].pre)
{
int v=E[i].to;
if(v!=fa)
dfs(v,now);
}
OUT[now]=total;
}
void init()
{
total=0;
MM(in,0);
MM(OUT,0);
MM(arr,0);
MM(ref,0);
MM(head,-1);
cnt=0;
}
void pushup(int k)
{
T[k].maxm=max(T[LC(k)].maxm,T[RC(k)].maxm);
T[k].minm=min(T[LC(k)].minm,T[RC(k)].minm);
}
void pushdown(int k)
{
if(T[k].add)
{
T[LC(k)].add+=T[k].add;
T[RC(k)].add+=T[k].add;
T[LC(k)].maxm+=T[k].add;
T[RC(k)].maxm+=T[k].add;
T[LC(k)].minm+=T[k].add;
T[RC(k)].minm+=T[k].add;
T[k].add=0;
}
}
void build(int k,int l,int r)
{
T[k].l=l;
T[k].r=r;
T[k].mid=MID(l,r);
T[k].add=T[k].maxm=T[k].minm=0;
if(l==r)
{
T[k].maxm=T[k].minm=arr[ref[l]];//初值的处理
return ;
}
build(LC(k),l,T[k].mid);
build(RC(k),T[k].mid+1,r);
pushup(k);
}
void update(int k,int l,int r,int val)
{
if(r<T[k].l||l>T[k].r)
return ;
if(l<=T[k].l&&r>=T[k].r)
{
T[k].add+=val;
T[k].maxm+=val;
T[k].minm+=val;
}
else
{
pushdown(k);
update(LC(k),l,r,val);
update(RC(k),l,r,val);
pushup(k);
}
}
int mmquery(int k,int l,int r)
{
if(l<=T[k].l&&r>=T[k].r)
return T[k].maxm;
pushdown(k);
if(r<=T[k].mid)
return mmquery(LC(k),l,r);
else if(l>T[k].mid)
return mmquery(RC(k),l,r);
else
return max(mmquery(LC(k),l,T[k].mid),mmquery(RC(k),T[k].mid+1,r));
}
int mnquery(int k,int l,int r)
{
if(l<=T[k].l&&r>=T[k].r)
return T[k].minm;
pushdown(k);
if(r<=T[k].mid)
return mnquery(LC(k),l,r);
else if(l>T[k].mid)
return mnquery(RC(k),l,r);
else
return min(mnquery(LC(k),l,T[k].mid),mnquery(RC(k),T[k].mid+1,r));
}
int Scan()
{
int res=0,ch,flag=0;
if((ch=getchar())=='-')
flag=1;
else if(ch>='0'&&ch<='9')
res=ch-'0';
while((ch=getchar())>='0'&&ch<='9')
res=res*10+ch-'0';
return flag?-res:res;
}
int main(void)
{
int tcase, i, x, q, val;
char ops[3];
scanf("%d",&tcase);
while (tcase--)
{
init();
scanf("%d",&n);
for (i=1; i<n; ++i)
{
x=Scan();
add(x,i+1);
}
for (i=1; i<=n; ++i)
arr[i]=Scan();
dfs(1,-1);
build(1,1,n);
scanf("%d",&q);
while (q--)
{
scanf("%s",ops);
if(ops[0]=='R')
{
scanf("%d%d",&x,&val);
update(1,in[x],OUT[x],val);
}
else
{
scanf("%d",&x);
int m=mmquery(1,in[x],OUT[x]),n=mnquery(1,in[x],OUT[x]);
printf("%d\n",m-n);
}
}
}
return 0;
}
另一种写法:
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define MM(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=1000010;
struct info
{
int to;
int pre;
}E[N];
int head[N],cnt;
int total,in[N],out[N];
int arr[N];
int n;
struct seg
{
int l,mid,r;
int maxm,minm,add;
};
seg T[N<<2];
void add(int s,int t)
{
E[cnt].to=t;
E[cnt].pre=head[s];
head[s]=cnt++;
}
void dfs(int now,int fa)
{
in[now]=++total;
for (int i=head[now]; ~i; i=E[i].pre)
{
int v=E[i].to;
if(v!=fa)
dfs(v,now);
}
out[now]=total;
}
void init()
{
MM(head,-1);
cnt=0;
total=0;
MM(in,0);
MM(out,0);
MM(arr,0);
}
void pushup(int k)
{
T[k].maxm=max(T[LC(k)].maxm,T[RC(k)].maxm);
T[k].minm=min(T[LC(k)].minm,T[RC(k)].minm);
}
void pushdown(int k)
{
if(!T[k].add)
return ;
T[LC(k)].add+=T[k].add;
T[RC(k)].add+=T[k].add;
T[LC(k)].maxm+=T[k].add;
T[RC(k)].maxm+=T[k].add;
T[LC(k)].minm+=T[k].add;
T[RC(k)].minm+=T[k].add;
T[k].add=0;
}
void build(int k,int l,int r)
{
T[k].l=l;
T[k].r=r;
T[k].mid=MID(l,r);
T[k].add=T[k].maxm=T[k].minm=0;
if(l==r)
return ;
build(LC(k),l,T[k].mid);
build(RC(k),T[k].mid+1,r);
}
void update(int k,int l,int r,int val)
{
if(r<T[k].l||l>T[k].r)
return ;
if(l<=T[k].l&&r>=T[k].r)
{
T[k].add+=val;
T[k].maxm+=val;
T[k].minm+=val;
}
else
{
pushdown(k);
update(LC(k),l,r,val);
update(RC(k),l,r,val);
pushup(k);
}
}
int mmquery(int k,int l,int r)
{
if(l<=T[k].l&&r>=T[k].r)
return T[k].maxm;
pushdown(k);
if(r<=T[k].mid)
return mmquery(LC(k),l,r);
else if(l>T[k].mid)
return mmquery(RC(k),l,r);
else
return max(mmquery(LC(k),l,T[k].mid),mmquery(RC(k),T[k].mid+1,r));
}
int mnquery(int k,int l,int r)
{
if(l<=T[k].l&&r>=T[k].r)
return T[k].minm;
pushdown(k);
if(r<=T[k].mid)
return mnquery(LC(k),l,r);
else if(l>T[k].mid)
return mnquery(RC(k),l,r);
else
return min(mnquery(LC(k),l,T[k].mid),mnquery(RC(k),T[k].mid+1,r));
}
int main(void)
{
int tcase, i, j, x, y, q, val;
char ops[3];
scanf("%d",&tcase);
while (tcase--)
{
init();
scanf("%d",&n);
for (i=1; i<n; ++i)
{
scanf("%d",&x);
y=i+1;
add(x,y);
}
for (i=1; i<=n; ++i)
scanf("%d",&arr[i]);
dfs(1,-1);
build(1,1,n);
for (i=1; i<=n; ++i)
update(1,in[i],in[i],arr[i]);//对于初值的处理
scanf("%d",&q);
while (q--)
{
scanf("%s",ops);
if(ops[0]=='R')
{
scanf("%d%d",&x,&val);
update(1,in[x],out[x],val);
}
else
{
scanf("%d",&x);
int m=mmquery(1,in[x],out[x]),n=mnquery(1,in[x],out[x]);
printf("%d\n",m-n);
}
}
}
return 0;
}
NBOJv2 1034 Salary Inequity(DFS序+线段树区间更新区间(最值)查询)的更多相关文章
- POJ.3321 Apple Tree ( DFS序 线段树 单点更新 区间求和)
POJ.3321 Apple Tree ( DFS序 线段树 单点更新 区间求和) 题意分析 卡卡屋前有一株苹果树,每年秋天,树上长了许多苹果.卡卡很喜欢苹果.树上有N个节点,卡卡给他们编号1到N,根 ...
- Educational Codeforces Round 6 E dfs序+线段树
题意:给出一颗有根树的构造和一开始每个点的颜色 有两种操作 1 : 给定点的子树群体涂色 2 : 求给定点的子树中有多少种颜色 比较容易想到dfs序+线段树去做 dfs序是很久以前看的bilibili ...
- 【BZOJ-3252】攻略 DFS序 + 线段树 + 贪心
3252: 攻略 Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 339 Solved: 130[Submit][Status][Discuss] D ...
- Codeforces 343D Water Tree(DFS序 + 线段树)
题目大概说给一棵树,进行以下3个操作:把某结点为根的子树中各个结点值设为1.把某结点以及其各个祖先值设为0.询问某结点的值. 对于第一个操作就是经典的DFS序+线段树了.而对于第二个操作,考虑再维护一 ...
- BZOJ2434 [Noi2011]阿狸的打字机(AC自动机 + fail树 + DFS序 + 线段树)
题目这么说的: 阿狸喜欢收藏各种稀奇古怪的东西,最近他淘到一台老式的打字机.打字机上只有28个按键,分别印有26个小写英文字母和'B'.'P'两个字母.经阿狸研究发现,这个打字机是这样工作的: 输入小 ...
- POJ 3321 DFS序+线段树
单点修改树中某个节点,查询子树的性质.DFS序 子树序列一定在父节点的DFS序列之内,所以可以用线段树维护. 1: /* 2: DFS序 +线段树 3: */ 4: 5: #include < ...
- 【XSY2667】摧毁图状树 贪心 堆 DFS序 线段树
题目大意 给你一棵有根树,有\(n\)个点.还有一个参数\(k\).你每次要删除一条长度为\(k\)(\(k\)个点)的祖先-后代链,问你最少几次删完.现在有\(q\)个询问,每次给你一个\(k\), ...
- F - Change FZU - 2277 (DFS序+线段树)
题目链接: F - Change FZU - 2277 题目大意: 题意: 给定一棵根为1, n个结点的树. 有q个操作,有两种不同的操作 (1) 1 v k x : a[v] += x, a[v ' ...
- BZOJ4551[Tjoi2016&Heoi2016]树——dfs序+线段树/树链剖分+线段树
题目描述 在2016年,佳媛姐姐刚刚学习了树,非常开心.现在他想解决这样一个问题:给定一颗有根树(根为1),有以下 两种操作:1. 标记操作:对某个结点打上标记(在最开始,只有结点1有标记,其他结点均 ...
- BZOJ1103 [POI2007]大都市meg dfs序 线段树
欢迎访问~原文出处——博客园-zhouzhendong 去博客园看该题解 题目传送门 - BZOJ1103 题意概括 一棵树上,一开始所有的边权值为1,我们要支持两种操作: 1. 修改某一条边的权值为 ...
随机推荐
- unsatisfied类型的异常
一般为libs包下arm64-v8a,armeabi,armeabi-v7a,mips,mips64,x86,x86_64等文件夹下的.so文件丢失. 一般情况,armeabi下的so文件需要拷贝一份 ...
- SELECT的解析顺序及慢查询优化
标准的 SQL 的解析顺序为:(1)FROM 子句, 组装来自不同数据源的数据(2)WHERE 子句, 基于指定的条件对记录进行筛选(3)GROUP BY 子句, 将数据划分为多个分组(4)使用聚合函 ...
- Android Bander设计与实现 - 设计篇
转自:http://blog.csdn.net/universus/article/details/6211589#t7 Binder Android IPC Linux 内核 驱动 摘要 Binde ...
- NVelocity模板引擎的使用
第一种使用方法直接赋值: VelocityEngine vltEngine = new VelocityEngine(); vltEngine.SetProperty(RuntimeConstants ...
- WPF中的VisualTreeHelper
VisualTreeHelper提供了一组WPF控件树模型,通过VisualTreeHelper可以遍历控件的树形结构,获取我们所需要的控件以及相应的属性: VisualTreeHelper提供了一下 ...
- Libsvm自定义核函数【转】
1. 使用libsvm工具箱时,可以指定使用工具箱自带的一些核函数(-t参数),主要有: -t kernel_type : set type of kernel function (default 2 ...
- VS2010编写动态链接库DLL及单元测试用例,调用DLL测试正确性
转自:http://blog.csdn.net/testcs_dn/article/details/27237509 本文将创建一个简单的动态链接库,并编写一个控制台应用程序使用该动态链接库,该动态链 ...
- android 定时器的实现
在Android上常用的定时器有两种,一种是Java.util.Timer,一种就是系统的AlarmService了. 实验1:使用Java.util.Timer. 在onStart()创创建Time ...
- LoadRunner中多值关联的3种处理方式
需求:通过关联取得的ParamName参数可能存在多个值,需要对每个ParamName参数值进行处理 脚本:可通过3种不同的实现方式,将每个参数值作为HTTP请求内容发出 web_reg_save_ ...
- unity button
#pragma strict var buttonTexture:Texture2D; private var str:String; private var frameTime:int; funct ...