hdu 3974 Assign the task

http://acm.hdu.edu.cn/showproblem.php?pid=3974

Assign the task

Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1113 Accepted Submission(s): 527



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

首先你要理解题意，这道题就是说在一个公司里，有领导和下属之分，给某一个员工派发任务之后，领导就会将任务派发给他的下属，知道任务被派发到最下层，这样一级一级的下发，问题是，对于任意一个员工的话，请你输出他现在在做的任务，如果没做就输出-1。这道题从题中需要注意的是从老板到下属是一个树，也就是说对于一个员工来说的话，他不可能有多个直接上司，当然可以有间接的，比如A是B的上司，C是A的上司，可以推出C也是B的上司，这一点儿很关键的，形成的是树，所以我们首先解决的是你的数据怎么来存的问题?我们可以用一维数组来存领导和下属的关系，来记住每个员工的直接领导是谁，这样其实这个数组就是一个树了，然后怎样解决任务的更新，因为题中说，对于每个员工当他接到新的任务时就会停下手中的任务而做新的任务，这样我想到用结构体了，用task来记住任务的名字，num来标记当前的任务数，num始终是需要更新的，这样才能保证输出的值是我们想要的当前的任务；

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <limits>
#include <queue>
#include <stack>
#include <vector>
#include <map>

using namespace std;

#define N 50005
#define INF 0xfffffff
#define PI acos (-1.0)
#define EPS 1e-8
#define Lson rt<<1, l, tree[rt].mid ()
#define Rson rt<<1|1, tree[rt].mid () + 1, r

struct node
{
    int task, num;
}stu
;

int main ()
{
    int t, a
, flag = 1;
    scanf ("%d", &t);
    while (t--)
    {
        int n, m;
        scanf ("%d", &n);
        memset (a, -1, sizeof (a));

        int x, y;
        for (int i=1; i<n; i++)
        {
            scanf ("%d%d", &x, &y);
            a[x] = y;
        }

        for (int i=1; i<=n; i++)
            stu[i].task = stu[i].num = -1;

        char s[10];
        int cnt = 0;
        printf ("Case #%d:\n", flag++);
        scanf ("%d", &m);

        while (m--)
        {
            scanf ("%s", s);
            if (s[0] == 'T')
            {
                scanf ("%d%d", &x, &y);
                stu[x].task = y;
                stu[x].num = ++cnt;
            }
            else
            {
                scanf ("%d", &x);
                y = stu[x].task;
                int tt = stu[x].num;

                while (x != -1)
                {
                    if (stu[x].num > tt)
                    {
                        y = stu[x].task;
                        tt = stu[x].num;
                    }
                    x = a[x];
                }
                printf ("%d\n", y);
            }
        }
    }
    return 0;
}