Assign the task HDU - 3974

题面

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.

法一：直接模拟多叉树+线段树

首先，读者可以画个树，自行标号dfs序，体会下dfs序的特点

不难发现，对于任意子树所含的结点编号是连续的

且该子树所含结点最小编号就是根结点编号

然后这个dfs序就是我们解决这道题的关键所在

按题意直接建好多叉树，标好dfs序，然后你会发现其实操作蛮简单的

查询员工，变成一层层去找对应子树根结点

#include <cstdio>
#include <set>
#include <cstring>

using namespace std;

const int maxn = 50005;

struct node{int dat, child, nxt, tag, boss, l, r;}tr[maxn];

int t, n, m, boss, tot, p[maxn], a, b;

void build(int p)
{
    tr[p].l = ++tot;
    for (int i = tr[p].child; i; i = tr[i].nxt)build(i);
    tr[p].r = tot;
}

inline void create(int Boss, int k)
{
    tr[k].boss = Boss;
    tr[k].nxt = tr[Boss].child;
    tr[Boss].child = k;
}

inline void push(int p)
{
    if (!tr[p].tag)return;
    for (int i = tr[p].child; i; i = tr[i].nxt)
        tr[i].dat = tr[i].tag = tr[p].tag;
    tr[p].tag = 0;
}

inline void update(int l, int r, int p)
{
    if (l == p) { tr[p].dat = tr[p].tag = r; return; }
    push(p);
    for (int i = tr[p].child; i;i=tr[i].nxt) 
        if (tr[i].l <= tr[l].l && tr[l].l <= tr[i].r) 
            update(l, r, i);
}


inline int query(int x, int p)
{
    if (x == p) return tr[p].dat?tr[p].dat:-1;
    push(p);
    for (int i = tr[p].child; i; i = tr[i].nxt) 
        if (tr[i].l <= tr[x].l && tr[x].l <= tr[i].r) 
            return query(x, i);
}

int main()
{
    scanf("%d", &t);
    for (int k = 1; k <= t; ++k)
    {
        scanf("%d", &n);
        printf("Case #%d:\n", k);
        memset(tr, 0, sizeof tr);
        for (int i = 1; i < n; ++i)
        {
            int a, b;
            scanf("%d%d", &a, &b);
            create(b, a);
        }
        for (int i = 1; i <= n; ++i)if (!tr[i].boss) { boss = i; break; }
        build(boss);
        scanf("%d", &m);
        for (int i = 1; i <= m; ++i)
        {
            char s[2];
            scanf("%s%d", s, &a);
            if (s[0] == 'C')printf("%d\n", query(a, boss));
            else { scanf("%d", &b); update(a, b, boss); }
        }
    }
}

　　

 

法二：并查集+模拟

每次查询，直接往上早，如果boss任务更新时间比当前任务晚，则更新当前任务

#include <cstdio>
using namespace std;
struct node
{int job, time;}a[50005];
int fa[50005], t, n, m, u, v, cnt;
char c[2];
int main()
{
    scanf("%d", &t);
    for (int cas = 1, cnt = 0; cas <= t; ++cas, cnt = 0)
    {
        scanf("%d",&n);
        for (int i = 1; i <= n; ++i) fa[i] = a[i].job = a[i].time = -1;
        for (int i = 1; i < n; ++i) scanf("%d%d",&u,&v), fa[u] = v;
        scanf("%d", &m);
        printf("Case #%d:", cas);
        while (m--)
        {
            int x;
            scanf("%s%d", c, x);
            if (c[0] == 'C')
            {
                int y = a[x].job, tmp = a[x].time;
                for (;x != -1;x=fa[x])
                    if (a[x].time > tmp) y = a[x].job, tmp = a[x].time;
                printf("%d\n", &y);
            }
            else
            {
                int y; scanf("%d", &y);
                a[x].job = y, a[x].time = ++cnt;
            }
        }
    }
    return 0;
}