J - Assign the task HDU 3974（线段树+树映射到区间 OR 时间戳 树的表示）

J - Assign the task
Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 3974
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 

大神们的博客基本是没有注释，也许是太简单了吧，可是我还是看了半天，尝试着理解了一些

做法是先把题目的树建立出来，然后DFS进行重新编号，使得老板和员工在一段连续的区间，这样就

转化成了线段树进行区间的更新，单点的查询，关键便是建立题目的树，以及遍历编号

对于很多人用的边的表示，我觉得是一种类似兄弟儿子表示法的方法（如果不对，还请指教）

#include<bits/stdc++.h>
using namespace std;
template<class T>inline T read(T&x)
{
    char c;
    while((c=getchar())<=32)if(c==EOF)return -1;
    bool ok=false;
    if(c=='-')ok=true,c=getchar();
    for(x=0; c>32; c=getchar())
        x=x*10+c-'0';
    if(ok)x=-x;
    return 1;
}
template<class T> inline T read_(T&x,T&y)
{
    return read(x)!=-1&&read(y)!=-1;
}
template<class T> inline T read__(T&x,T&y,T&z)
{
    return read(x)!=-1&&read(y)!=-1&&read(z)!=-1;
}
template<class T> inline void write(T x)
{
    if(x<0)putchar('-'),x=-x;
    if(x<10)putchar(x+'0');
    else write(x/10),putchar(x%10+'0');
}
template<class T>inline void writeln(T x)
{
    write(x);
    putchar('\n');
}
//-------ZCC IO template------
const int maxn=50010;
const double inf=999999999;
#define lson (rt<<1),L,M
#define rson (rt<<1|1),M+1,R
#define M ((L+R)>>1)
#define For(i,t,n) for(int i=(t);i<(n);i++)
typedef long long  LL;
typedef double DB;
#define bug printf("---\n");

struct Edge//
{
    int to,next;//树的边的表示，to表示边的下端点，next表示以边的上端点为公共点的相邻点
}edge[maxn];
bool vis[maxn];//标记那些点做过儿子，那么没有做过儿子的就是树根
int head[maxn],start[maxn],endd[maxn];//head[i]表示以i为结点的公共边的条的号码，start[i]到end[i]表示以结点i为老板的所有员工的编号
int tot,cnt;//tot表示边数，cnt表示DFS对各点的编码

 void init()
{
    tot=cnt=0;
    memset(head,-1,sizeof(head));
    memset(vis,false,sizeof(vis));
}
void addedge(int u,int v)//以边表示数
{
    edge[tot].to=v;
    edge[tot].next=head[u];
    head[u]=tot++;
}
void dfs(int u)
{
    start[u]=++cnt;//老板的起点编号
    for(int i=head[u];~i;i=edge[i].next)
        dfs(edge[i].to);
    endd[u]=cnt;//他最后的员工的编号
}

struct node
{
    int inv,val;
}p[maxn<<2];
void build(int rt,int L,int R)
{
    p[rt].val=-1;
    p[rt].inv=0;
    if(L==R)return ;
    build(lson);
    build(rson);
}
void update(int rt,int L,int R,int x,int y,int v)
{
    //printf("%d %d %d %d %d %d\n",rt,L,R,x,y,v);
    if(L==x&&y==R)
    {
        p[rt].val=v;
        p[rt].inv=1;
        return ;
    }
    if(p[rt].inv)
    {
        p[rt<<1].inv=p[rt<<1|1].inv=1;
        p[rt<<1].val=p[rt<<1|1].val=p[rt].val;
        p[rt].inv=0;
    }
    if(y<=M)
        update(lson,x,y,v);
    else if(x>M)
        update(rson,x,y,v);
    else
    {
        update(lson,x,M,v);
        update(rson,M+1,y,v);
    }
}
int query(int rt,int L,int R,int i)
{
    if(L==R)
    {
        return p[rt].val;
    }
    if(p[rt].inv)
    {
        p[rt<<1].inv=p[rt<<1|1].inv=1;
        p[rt<<1].val=p[rt<<1|1].val=p[rt].val;
        p[rt].inv=0;
    }
    if(i<=M)return query(lson,i);
    else return query(rson,i);
}

int main()
{
//    #ifndef ONLINE_JUDGE
//    freopen("in.txt","r",stdin);
//    #endif // ONLINE_JUDGE
    int n,m,i,j,k,t;
    int T;
    read(T);
    int cas=1;
    while(T--)
    {
        printf("Case #%d:\n",cas++);
        init();
        read(n);
        For(i,1,n)
        {
            int u,v;
            read_(u,v);
            vis[u]=true;
            addedge(v,u);
        }
        For(i,1,n+1)if(!vis[i])
        {
            dfs(i);break;
        }
        read(m);
        char op[2];
        build(1,1,cnt);
        while(m--)
        {
            scanf("%s",op);
            if(op[0]=='T')
            {
                int x,y;
                read_(x,y);
                update(1,1,cnt,start[x],endd[x],y);
            }
            else
            {
                int x;
                read(x);
                writeln(query(1,1,cnt,start[x]));
            }
        }
    }
    return 0;
}