Assign the task
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： 一个公司有n个职员，他们可能有自己的上司和下属，公司如果给其中一个职员分配任务，那么这个任务也会同时分配给他的下属，而且每个新任务分配时，此职员当前任务只能是新任务，现在要查询一些职员的当前任务是什么，若一直未分配任务则输出-1。
测试样例分析：
#include<stdio.h>
#define N 50010
int fa[N], vis[N], task[N];
//fa数组保存父节点，即上司
//task[i] 代表分配给i的任务为task[i]
//vis数组用于记录任务分配的优先级
int find(int x){
int r = x, v = vis[x], Working_task = task[x];
//v是任务序号，ans是任务
while(fa[r] != r){
if(v<vis[fa[r]]){ //如果找到更靠后的序号，那就更新任务，及其序号
Working_task = task[fa[r]];
v = vis[fa[r]];
}
r = fa[r]; //一直查找其父节点，直自其根节点是本身
}
return Working_task;
}
int main(){
int T, num; //num记录的是分配任务的序号，那么当前任务一定是最后分配的
int num_Test = 0,num_Message;
scanf("%d",&T);//T组测试样例
while(T--){
num_Test++;
num = 1;
int n;
scanf("%d",&n);//n个人
//初始化
for(int i = 0;i<=n;i++){
fa[i] = i;//每个点都是独立的
vis[i] = 0;//记录访问i的次数
task[i] = -1; //记录i当前任务
}
//输入
for(int i = 1;i<n;i++){//输入关系
int a,b;
scanf("%d%d",&a,&b);//b是a的上司
fa[a] = b;
}
scanf("%d",&num_Message);//输入信息数量
printf("Case #%d:\n", num_Test);
while(num_Message--){
char operate[2];//操作符:C/T
scanf("%s", operate);
if(operate[0] == 'T'){ //T为输入数据
int a,b;
scanf("%d%d",&a,&b);
task[a] = b;
vis[a] = num++;//每次分配任务，要记下序号,序号越往后，代表越新的
}
else if(operate[0] == 'C'){
int a;
scanf("%d",&a);
printf("%d\n", find(a));
}
}
}
return 0;
}
网上好多用的是dfs+线段树，奈何我看了一上午，还是没看懂(T^T)，待我学了线段树，再回来看看吧
#include<stdio.h>
#include<string.h>
#include<queue>
#define inf 0x3f3f3f3f
#define MAXN 50005
#define ll long long
using namespace std;
int n,m;
int ord[MAXN*4],root,tot,tmp[MAXN*4],ed[MAXN*4];
int laz[MAXN*4];
bool hasIn[MAXN];
vector<int> vec[MAXN];
/*
//假设root --> a --> b
+->c
dfs(root){
tmp[root] = 1; ord[1] = root;
for(int i = 0;i<2;i++){ //vec[root].size() = 2
//第一轮
int v = a; //int v = vec[root][0];
dfs(a){ //dfs(a);
tmp[a] = 2; ord[2] = a;
for(int i = 0;i<1;i++){//vec[a].size() = 1
int v = c;//int v=vec[a][0];
dfs(c){
tmp[c] = 3; ord[3] = c;
for{...} //不符合直接跳过
ed[c] = 3;//tot = 3
}
}
}
//第二轮
int v = b; //int v = vec[root][1];
dfs(b){
tmp[b] = 4; ord[4] = b;
for{...} //不符合直接跳过
ed[b] = 4;
}
}
ed[root] = 4
}
*/
void dfs(int u){//开始传进根节点
tmp[u]=++tot;
ord[tot]=u;
for(int i=0;i<vec[u].size();i++){
int v=vec[u][i];
dfs(v);
}
ed[u]=tot;
}
void pushdown(int v)
{
laz[v*2]=laz[v];
laz[v*2+1]=laz[v];
laz[v]=-1;
}
void updata(int v,int L,int R,int ql,int qr,int d)
{
if(ql<=L && R<=qr)
{
laz[v]=d;
return;
}
if(laz[v]!=-1) pushdown(v);
int mid=(L+R)/2;
if(ql<=mid) updata(v*2,L,mid,ql,qr,d);
if(qr>mid) updata(v*2+1,mid+1,R,ql,qr,d);
}
/*
//还是假设 root --> a --> b ,假设查询 b节点
+-> c
int query(1,1,4,4){ //v = 1;L = 1;R = 4;q = tmp[b]=4;
if(1==4){...} //不满足
if(laz[1]!=-1){...} //laz[1] = -1;不满足
int mid = 1; //(1+1)/2
if(4<=1){...} //不满足
else
return query(3,2,4,4){
}
}
*/
int query(int v,int L,int R,int q)
{
if(L==R)
return laz[v];
if(laz[v]!=-1)
pushdown(v);
int mid=(L+R)/2;
if(q<=mid)
return query(v*2,L,mid,q);
else
return query(v*2+1,mid+1,R,q);
}
int main(){
int t; //t组测试样例
scanf("%d",&t);
for(int num_t=1;num_t<=t;num_t++){
printf("Case #%d:\n",num_t);
scanf("%d",&n);//输入总节点数
//初始化
for(int i = 1;i<=n;i++)
vec[i].clear();
memset(hasIn,false,sizeof(hasIn));
memset(tmp,0,sizeof tmp);
memset(laz,-1,sizeof laz);
tot=0;
for(int i = 1;i<n;i++){
int u,v;//用于表示输入的两个数
scanf("%d%d",&u,&v);// v --> u
vec[v].push_back(u);//vec[v]存放所有v所指向的节点
hasIn[u]=true;
}
for(int i = 1;i<=n;i++)//扫描所有节点,找出没有入度的节点
if(!hasIn[i]){//没有入度，代表是根节点
root=i;
break;
}
dfs(root);//深搜根节点
char operate[2];//操作符号:C/T
scanf("%d",&m);//m个信息
while(m--){
scanf("%s",operate);
if(operate[0]=='C'){
//C为查询，则输入查询节点u
int u;
scanf("%d",&u);
printf("%d\n",query(1,1,tot,tmp[u]));
}
else if(operate[0]=='T'){
int u,d;
scanf("%d%d",&u,&d);
updata(1,1,tot,tmp[u],ed[u],d);
}
}
}
return 0;
}
京公网安备 11010502036488号