hdu 2275

Kiki & Little Kiki 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1027    Accepted Submission(s): 335

Problem Description

Kiki is considered as a smart girl in HDU, many boys fall in love with her! Now, kiki will finish her education, and leave school, what a pity! One day, zjt meets a girl, who is like kiki very much, in the campus, and calls her little kiki. Now, little kiki want to design a container, which has two kinds of operation, push operation, and pop operation.
Push M:
  Push the integer M into the container.
Pop M:
  Find the maximal integer, which is not bigger than M, in the container. And pop it from the container. Specially, for all pop operations, M[i] is no bigger than M[i+1].
Although she is as smart as kiki, she still can't solve this problem! zjt is so busy, he let you to help little kiki to solve the problem. Can you solve the problem?

 

Input

The input contains one or more data sets. At first line of each input data set is an integer N (1<= N <= 100000) indicate the number of operations. Then N lines follows, each line contains a word (“Push” or “Pop”) and an integer M. The word “Push” means a push operation, while “Pop” means a pop operation. You may assume all the numbers in the input file will be in the range of 32-bit integer.

 

Output

For each pop operation, you should print the integer satisfy the condition. If there is no integer to pop, please print “No Element!”. Please print a blank line after each data set.

 

Sample Input

9
Push 10
Push 20
Pop 2
Pop 10
Push 5
Push 2
Pop 10
Pop 11
Pop 19
3
Push 2
Push 5
Pop 2

 

Sample Output

No Element!
10
5
2
No Element!

2

 

////////////////////////////////////////

//2275
//弄两个优先队列  
//一个从大到小，一个从小到大  
//我们只需要每次看推入的这个和上一个输出的谁大？如果比上个输出的小，推入1队列。否则推入2队列。
//然后后面发现multiset这个集合，更简单了


#include <iostream>
#include <string>
#include <set>
using namespace std;


int main()
{
multiset<int> st;
multiset<int>::iterator it;
    string str;
    int num, n;
    while(scanf("%d",&n) != EOF)
    {
        st.clear();  //清零
        while(n--)
        {
            cin>>str>>num;
            if (str[1] == 'u')  
            {  
                st.insert(num);  
            }  
            else  
            {  
                it = st.upper_bound(num); //upper_bound()返回的是在前闭后开区间查找的关键字的上界
                if (it == st.begin())
                {  
                    printf("No Element!\n");
                }     
                else  
                {  
                    it--;  
                    printf("%d\n", *it);  
                    st.erase(it);  
                }  
            }  
        }
        printf("\n");
    }
    return 0;
}