Assign the task
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 780 Accepted Submission(s): 392
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.
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)
1 5 4 3 3 2 1 3 5 2 5 C 3 T 2 1 C 3 T 3 2 C 3
Case #1: -1 1 2
先dfs改时间戳,那么子树区间就连在一起了,然后就是区间更新单点查询
/*************************************************************************
> File Name: hdu3974.cpp
> Author: ALex
> Mail: 405045132@qq.com
> Created Time: 2015年01月11日 星期日 20时50分13秒
************************************************************************/
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 50110;
struct node
{
int l, r;
int add;
int job;
}tree[N << 2];
vector <int> edge[N];
int s[N], e[N], cnt;
int deg[N];
void dfs (int u)
{
int m = edge[u].size();
s[u] = ++cnt;
for (int i = 0; i < m; ++i)
{
int v = edge[u][i];
dfs (v);
}
e[u] = cnt;
}
void pushdown (int p)
{
if (tree[p].add != -1)
{
tree[p << 1].add = tree[p << 1 | 1].add = tree[p].add;
tree[p << 1].job = tree[p << 1 | 1].job = tree[p].add;
tree[p].add = -1;
}
}
void build (int p, int l, int r)
{
tree[p].l = l;
tree[p].r = r;
tree[p].add = -1;
tree[p].job = -1;
if (l == r)
{
return;
}
int mid = (l + r) >> 1;
build (p << 1, l, mid);
build (p << 1 | 1, mid + 1, r);
}
void update (int p, int l, int r, int val)
{
if (l == tree[p].l && r == tree[p].r)
{
tree[p].job = val;
tree[p].add = val;
return;
}
pushdown (p);
int mid = (tree[p].l + tree[p].r) >> 1;
if (r <= mid)
{
update (p << 1, l, r, val);
}
else if (l > mid)
{
update (p << 1 | 1, l, r, val);
}
else
{
update (p << 1, l, mid, val);
update (p << 1 | 1, mid + 1, r, val);
}
}
int query (int p, int pos)
{
if (tree[p].l == tree[p].r)
{
return tree[p].job;
}
pushdown (p);
int mid = (tree[p].l + tree[p].r) >> 1;
if (pos <= mid)
{
return query (p << 1, pos);
}
else
{
return query (p << 1 | 1, pos);
}
}
int main()
{
int t, n, m;
int icase = 1;
char op[10];
int x, y;
scanf("%d", &t);
while (t--)
{
scanf("%d", &n);
memset (deg, 0, sizeof(deg));
for (int i = 0; i <= n; ++i)
{
edge[i].clear();
}
int u, v;
for (int i = 1; i <= n - 1; ++i)
{
scanf("%d%d", &u, &v);
edge[v].push_back(u);
deg[u]++;
}
int root;
for (int i = 1; i <= n; ++i)
{
if (!deg[i])
{
root = i;
break;
}
}
cnt = 0;
dfs (root);
build (1, 1, cnt);
scanf("%d", &m);
printf("Case #%d:\n", icase++);
while (m--)
{
scanf("%s", op);
if (op[0] == 'C')
{
scanf("%d", &x);
printf("%d\n", query(1, s[x]));
}
else
{
scanf("%d%d", &x, &y);
int l = s[x];
int r = e[x];
update (1, l, r, y);
}
}
}
return 0;
}