HDU 3974 Assign the task (DFS + 线段树)

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Assign the task

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3816    Accepted Submission(s): 1581

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

【题意】

给 个树 关系图，   每次 修改和查询，  查询 当前点 的状态，    实质为区间修改 单点查询

【思路】

DFS 先将树的关系 列出， 然后 区间线段树

【代码实现】

//#include <bits/stdc++.h>
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <math.h>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <vector>
#define mem(a,b) memset(a,b,sizeof(a))
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define FIN      freopen("input.txt","r",stdin)
#define FOUT     freopen("output.txt","w",stdout)
#define S1(n)    scanf("%d",&n)
#define SL1(n)   scanf("%I64d",&n)
#define S2(n,m)  scanf("%d%d",&n,&m)
#define SL2(n,m)  scanf("%I64d%I64d",&n,&m)
#define Pr(n)     printf("%d\n",n)
#define lson rt << 1, l, mid
#define rson rt << 1|1, mid + 1, r

using namespace std;
typedef long long ll;
const double PI=acos(-1);
const int INF=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e6+5;
const int MAXN=50010;
const int MOD=1e9+7;
const int mod=1e9+7;

int add[MAXN<<2],sum[MAXN<<2];
int head[MAXN],flag[MAXN];
int le[MAXN],ri[MAXN];
int cot,key;

struct  node
{
	int v,next;
}edge[MAXN];

void init()
{
	cot=0;
	mem(head,-1);
}

void add_edge(int u,int v)
{
	edge[cot].v=v;
	edge[cot].next=head[u];
	head[u]=cot++;
}

void PushDown(int rt)
{
	if(add[rt]!=-1)
	{
		sum[rt<<1]=add[rt];
		sum[rt<<1|1]=add[rt];
		add[rt<<1]=add[rt];
		add[rt<<1|1]=add[rt];
		add[rt]=-1;
	}
}
void dfs(int rt)
{
	le[rt]=++key;
	for(int i=head[rt];i!=-1;i=edge[i].next)
		dfs(edge[i].v);
	ri[rt]=key;
}
void build(int rt,int l,int r)
{
	sum[rt]=add[rt]=-1;
	if(l==r)
		return;
	int mid=(l+r)>>1;
	build(lson);
	build(rson);
}
void update(int rt,int l,int r,int L,int R,int C)
{
	if(L<=l&&R>=r)
	{
		sum[rt]=C;
		add[rt]=C;
		return;
	}
	PushDown(rt);
	int mid=(l+r)>>1;
	if(L<=mid) 	update(lson,L,R,C);
	if(R>mid)	update(rson,L,R,C);
}
int query(int rt,int l,int r,int L,int R)
{
	if(L<=l&&R>=r)
	{
		return sum[rt];
	}
	PushDown(rt);
	int mid=(l+r)>>1;
	int ans=0;
	if(L<=mid) ans+=query(lson,L,R);
	if(R>mid) ans+=query(rson,L,R);
	return ans;
}

int main()
{
	int t;
	int n,m,q,x,y;
	int cont=0;
	cin>>t;
	while(t--)
	{
		cin>>n;
		init();
		mem(flag,0);
		mem(ri,0);
		mem(le,0);
		for(int i=1;i<n;i++)
		{
			scanf("%d %d",&x,&y);
			add_edge(y,x);
			flag[x]=1;
		}
		key=0;
		for(int i=1;i<=n;i++)
		{
			if(!flag[i])
			{
				dfs(i);
				break;
			}
		}
		build(1,1,key);
		cin>>m;
		printf("Case #%d:\n",++cont);
		while(m--)
		{
			char op[5];
			scanf("%s",op);
			if(op[0]=='C')
			{
				scanf("%d",&x);
				printf("%d\n",query(1,1,key,le[x],le[x]));
			}
			else
			{
				scanf("%d %d",&x,&y);
				//printf("%d %d\n",le[x],ri[x]);
				update(1,1,key,le[x],ri[x],y);
			}

		}

	}
	return 0;
}

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