HDU

Assign the task

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

 解题思路：DFS把树转换成区间，然后可以通过线段树去维护这些区间，从而变为线段树问题。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <set>#include <vector>#include <map>#include <stack>#include <set>#include <algorithm>using namespace std;typedef long long ll;const int maxn=50005;int tree[maxn<<2];//线段树数组，看你要存什么void build(int l,int r,int rt){    tree[rt]=-1;    if(l==r){        return;    }    int m=(l+r)>>1;//>>1等于/2    //递归建树    build(l,m,rt<<1);    build(m+1,r,rt<<1|1);}//下推函数，ln,rn为左子树，右子树的数字数量。void pushdown(int rt){    //修改子节点的值    tree[rt<<1]=tree[rt<<1|1]=tree[rt];    //清除标记    tree[rt]=-1;}//区间修改，比点修改多了个R参数，这里是区间+C，与查询同理，在查询时修改void update(int L,int R,int C, int l, int r, int rt){    if(L<=l&&r<=R){//如果当前区间在查询区间内，直接对其进行修改        tree[rt]=C;//C*当前区间含有的数字个数        return;    }    int m=(l+r)>>1;    if(tree[rt]!=-1)        pushdown(rt);    if(L <= m) //与查询同理        update(L,R,C,l,m,rt<<1);    if(R >  m)        update(L,R,C,m+1,r,rt<<1|1);}//查询，这里为求最值,LR代表要查询的区间,lr代表当前区间，rt表示当前节点在数组中的实际位置int query(int L,int R,int l,int r,int rt){    if(l==r)        return tree[rt];    int m=(l+r)>>1;    //每次查询都下推标记，保证值正确    if(tree[rt]!=-1)    pushdown(rt);    if(L<=m)    return query(L,R,l,m,rt<<1);    else    return query(L,R,m+1,r,rt<<1|1);}struct node{    int parent=0;    vector<int> childs;};int li[maxn],ri[maxn];int len=1;void dfs(int cn,node* tn){    li[cn]=len;    for(int i=0;i<tn[cn].childs.size();i++){        dfs(tn[cn].childs[i],tn);    }    ri[cn]=len++;}int main(){    int t;    scanf("%d",&t);    for(int qqq=1;qqq<=t;qqq++){        printf("Case #%d:\n",qqq);    int n,q;    scanf("%d",&n);    len=1;    node* tn=new node[n+1];    for(int i=1;i<n;i++){        int a,b;        scanf("%d%d",&a,&b);        tn[b].childs.push_back(a);        tn[a].parent=b;    }    for(int i=1;i<=n;i++){        if(tn[i].parent==0){            dfs(i,tn);            break;        }    }    build(1,len,1);    scanf("%d",&q);    for(int i=0;i<q;i++){        char c[10];        scanf("%s",c);        if(c[0]=='C'){            int temp;            scanf("%d",&temp);            printf("%d\n",query(ri[temp],ri[temp],1,len,1));        }        else{            int temp,col;            scanf("%d%d",&temp,&col);            update(li[temp],ri[temp],col,1,len,1);        }    }    }    return 0;}