HDU3974 Assign the task
Assign the task
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5032 Accepted Submission(s): 1966
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.
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)
5
4 3
3 2
1 3
5 2
5
C 3
T 2 1
C 3
T 3 2
C 3
-1
1
2
#include<iostream>
#include<vector>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 5e5+8;
int First[maxn<<1],LAST[maxn<<1],wa;
vector<int>u[maxn];
bool book[maxn];
struct node
{
int l,r;
int num;
bool lazy;
}tree[maxn<<2];
void dfs(int t)
{
int siz=u[t].size();
First[t]=++wa;
for(int i=0;i<siz;i++){
dfs(u[t][i]);
}
LAST[t]=++wa;
} void build(int t,int l,int r)
{
tree[t].l=l;tree[t].r=r;
tree[t].num=-1;tree[t].lazy=false;
if(l==r){return;} int mid=(l+r)>>1; build(t<<1,l,mid);
build((t<<1)|1,mid+1,r);
} void push_down(int t)
{
tree[t<<1].num=tree[t].num;
if(tree[t<<1].l!=tree[t<<1].r){tree[t<<1].lazy=true;}
tree[(t<<1)|1].num=tree[t].num;
if(tree[(t<<1)|1].l!=tree[(t<<1)|1].r){tree[(t<<1)|1].lazy=true;}
tree[t].lazy=false;
} void update(int t,int l,int r,int x)
{
if(tree[t].lazy)push_down(t);
if(tree[t].l==l&&tree[t].r==r){
if(l!=r)tree[t].lazy=true;
tree[t].num=x;return ;
} int mid=(tree[t].l+tree[t].r)>>1;
if(r<=mid){update(t<<1,l,r,x);}
else if(l>mid){update((t<<1)|1,l,r,x);}
else {
update((t<<1),l,mid,x);
update((t<<1)|1,mid+1,r,x);
}
} int query(int t,int x)
{
// cout<<t<<" "<<x<<" "<<tree[t].lazy<<endl;
if(tree[t].lazy)push_down(t);
if(tree[t].l==tree[t].r){
if(tree[t].l!=x){return -1;}
return tree[t].num;
} int mid=(tree[t].l+tree[t].r)>>1;
if(x<=mid){return query(t<<1,x);}
else if(x>mid){return query((t<<1)|1,x);}
} int main()
{
int T,n;
scanf("%d",&T);
int re=0;
while(T--){
re++;
printf("Case #%d:\n",re);
wa=0;
memset(book,0,sizeof(book));
memset(First,0,sizeof(First));
memset(LAST,0,sizeof(LAST));
scanf("%d",&n);
for(int i=0;i<=n+1;i++){
u[i].clear();
}
int x,y;
for(int i=1;i<n;i++){
scanf("%d%d",&x,&y);
u[y].push_back(x);
book[x]++;
}
for(int i=1;i<=n;i++){
if(!book[i]){dfs(i);}
} build(1,1,n*2); scanf("%d",&n);
char p[5];
for(int i=1;i<=n;i++){
getchar();
scanf("%s",p);
if(p[0]=='C'){
scanf("%d",&x);
printf("%d\n",query(1,First[x]));
}
else if(p[0]=='T'){
scanf("%d%d",&x,&y);
update(1,First[x],LAST[x],y);
} }
}
return 0;
}
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