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Time Limit : 15000/5000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 4 Accepted Submission(s) : 2
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Problem Description
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
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
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
Source
//查找的时候只需查找最近更新的上司(这里要查遍某个节点所有的上司,找出谁是最近更新的)
#include<iostream>
#include<cstring>
using namespace std;
const int maxn=50005;
struct node{
int time,key; //设置一个time来判断是否是最近更新的数据
}node[maxn];
int parent[maxn];
int main(){
int t,cnt,n,m,u,v;
char str[5];
scanf("%d",&t);
for(cnt=1;cnt<=t;cnt++){
memset(parent,-1,sizeof(parent));
scanf("%d",&n);
for(int i=1;i<=n-1;i++){
scanf("%d%d",&u,&v);
parent[u]=v;
}
for(int i=1;i<=n;i++){
node[i].time=-1;
node[i].key=-1;
}
scanf("%d",&m);
printf("Case #%d:\n",cnt);
int x,k,T=0;
while(m--){
scanf("%s",str);
if(str[0]=='T'){
scanf("%d%d",&x,&k);
node[x].key=k;
node[x].time=++T; //注意这一步
}
else{
scanf("%d",&x);
k=node[x].key;
int time=node[x].time;
while(x!=-1){ //查找最近更新的上司
if(node[x].time>time){
k=node[x].key;
time=node[x].time;
}
x=parent[x];
}
printf("%d\n",k);
}
}
}
//system("pause");
return 0;
}