HDU-3974 Assign the task (dfs序+线段树区间修改点查询)

Assign the task

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3895    Accepted Submission(s): 1619

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

#include <bits/stdc++.h>using namespace std;const int maxn = 50001;int l[maxn], r[maxn], tot, c[maxn << 2], pre[maxn];vector<vector<int> > g(maxn);void dfs(int x, int fa){l[x] = ++tot;int cur;for(int i = 0; i < g[x].size(); ++i){cur = g[x][i];if(cur == fa) continue;dfs(cur, x);}r[x] = tot;}void build(int o, int l, int r){c[o] = -1;if(l == r) return;int mid = l + r >> 1;build(o << 1, l, mid);build(o << 1 | 1, mid + 1, r);}void pushdown(int o, int l, int r){if(l == r || c[o] == -1) return;c[o << 1] = c[o << 1 | 1] = c[o];c[o] = -1;}int query(int o, int l, int r, int id){pushdown(o, l, r);if(l == r) return c[o];int mid = l + r >> 1;if(id <= mid) return query(o << 1, l, mid, id);else return query(o << 1 | 1, mid + 1, r, id);}void update(int o, int l, int r, int L, int R, int v){pushdown(o, l, r);if(l >= L && r <= R){c[o] = v;return;}int mid = l + r >> 1;if(mid >= L) update(o << 1, l, mid, L, R, v);if(mid < R) update(o << 1 | 1, mid + 1, r, L, R, v);}int main(){int T, casenum = 1;scanf("%d", &T);while(T--){printf("Case #%d:\n", casenum++);memset(pre, -1, sizeof(pre));int u, v, n, q, rt;char s[3];scanf("%d", &n);for(int i = 1; i <= n; ++i){g[i].clear();}for(int i = 1; i < n; ++i){scanf("%d %d", &u, &v);g[v].push_back(u);pre[u] = v;}for(int i = 1; i <= n; ++i){if(pre[i] == -1){rt = i;break;}}tot = 0;dfs(rt, -1);build(1, 1, n);scanf("%d", &q);while(q--){scanf("%s", s);if(s[0] == 'T'){scanf("%d %d", &u, &v);update(1, 1, n, l[u], r[u], v);}else{scanf("%d", &u);printf("%d\n", query(1, 1, n, l[u]));}}}}/*题意：一棵树，5e4个节点，5e4次操作，每次操作要么讲以某节点为根的子树整体修改为某个数字，要么查询某个节点当前数字是多少。起始都是-1。思路：dfs序处理一下，给树上节点编号，某棵子树上所有点的编号的会在一个范围内，对应到线段树上区间维护即可。线段树上区间修改点查询就是基本操作了。*/