In this cafeteria, the N tables are all ordered in one line, where table number 1 is the closest to the window and table number N is the closest to the door.
Each time a group of X people enter the cafeteria, one of the cafeteria staff escorts the guests to the table with the smallest number of chairs greater than or equal to X chairs among all available tables. If there’s more than one such table, the guests are escorted to the table that is closest to the window. If there isn't any suitable table available, the group will leave the cafeteria immediately. A table is considered available if no one sits around it.
Eyad likes to help others and also likes to prove that he has learned something from the training of Master Hasan. Therefore, he decided to write a program that helps the cafeteria staff choose the right table for a newly arriving group.
Given the log file of who entered and who left the cafeteria, find the table at which a given group should sit.
Input
The first line of input contains two integers N and Q (1 ≤ N, Q ≤ 105), the number of tables in the cafeteria and the number of events in the log file.
The second line contains N integers, each represents the size of a table (number of chairs around it). The tables are given in order from 1 to N. The size of each table is at least 1 and at most 105.
Each of the following Q lines describes an event in one of the following formats:
- in X: means a group of X (1 ≤ X ≤ 105) people entered the cafeteria.
- out T: means the group on table number T (1 ≤ T ≤ N) just left the cafeteria, the table is now available. It is guaranteed that a group was sitting on this table (input is valid).
Initially, all tables are empty, and the log lists the events in the same order they occurred (in chronological order).
Output
For each event of the first type, print the number of the table on which the coming group should sit. If for any event a group cannot be escorted to any table, print - 1 for that event.
Example
4 7
1 2 6 7
in 4
in 1
in 3
in 5
out 1
out 4
in 7
3
1
4
-1
做這題之前先來普及下 set<pair<int,int> > 的用法 注意：set<pair<int,int> >; // 最后的> > 是 >空格> 否則會報錯
其次
set<pair<int,int> >在頭文件<set>中，並且要加上using namespace std；
<vector> 頭文件 // iterator 迭代器迭代程序
set<pair<int,int> > // 是一種類，注意：定義的時候右邊的兩個> >要空一格。
在頭文件<set>中
set<pair<int,int> > s // 聲明 s s是類名，
set<pair<int,int> >::iterator it; // iterator 迭代器迭代程序
lower_bound(key_value) 返回第一個大於等於key_value的定位器
upper_bound(key_value)，返回最后一個大於等於key_value的定位器
假若上面查找的沒有符合條件的，那么他們的值等於 s.end（）
s.erase(it); 釋放掉it的值
s.insert(make_pair(a[i],i)) 插入
it=s.lower_bound(make_pair(t,0));//尋找答案
ac代碼：
#include<iostream>
#include<set>
#include<vector>
using namespace std;
const int maxn=1e5+10;
int n,q,a[maxn];
set<pair<int,int> > s; //最后兩個> > 是 >空格> 否則會報錯
set<pair<int,int> >::iterator it; // iterator 迭代器迭代程序
int main() //set默認的比較規則先按照first比較，如果first相同，再按照second 比較。
{
cin>>n>>q;
int i,x;
char str[10];
for (i=1;i<=n;i++){
cin>>a[i];
s.insert(make_pair(a[i],i)); //插入椅子數及其桌號
}
while (q--){
cin>>str>>x;
if (str[0]=='i'){
it=s.lower_bound(make_pair(x,0)); //lower_bound(key_value) 返回第一個大於等於key_value的定位器
if (it==s.end()){ //假若上面查找的沒有符合條件的，那么他們的值等於 s.end（）
cout<<"-1\n";
continue;
}
else{
cout<<it->second<<endl;
s.erase(it);//釋放掉了桌號，相當於標記該桌號不能再用
}
}
else{
s.insert(make_pair(a[x],x));
}
}
return 0;
}