HDU 5347 多校赛1007 MZL's simple problem

Problem Description

   
   
   
   

    
    
    
    A simple problem
Problem Description
You have a multiple set,and now there are three kinds of operations:
1 x : add number x to set
2 : delete the minimum number (if the set is empty now,then ignore it)
3 : query the maximum number (if the set is empty now,the answer is 0)
   
   
   
   

 

Input

   
   
   
   

    
    
    
    The first line contains a number 
    
    
    
    
    
    
    
     N  (
    
    
    
    
    
    
    
     N≤106 ),representing the number of operations.
Next 
    
    
    
    
    
    
    
     N  line ,each line contains one or two numbers,describe one operation.
The number in this set is not greater than 
    
    
    
    
    
    
    
     109 .
   
   
   
   

 

Output

   
   
   
   

    
    
    
    For each operation 3,output a line representing the answer.
   
   
   
   

 

Sample Input

   
   
   
   

    
    
    
    6
1 2
1 3
3
1 3
1 4
3
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    3
4
   
   
   
   

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <stack>
#include <algorithm>
#define LL long long

using namespace std;

int main()
{
    priority_queue<int, vector<int>, greater<int> >q;
    int n;
    scanf("%d",&n);
    int ch,x;
    int MAX;
    while(n--)
    {
        scanf("%d",&ch);
        if(ch == 1)
        {
            scanf("%d",&x);
            if(q.size() == 0)
            {
                MAX = x;
                q.push(x);
            }
            else
            {
                if(MAX < x)
                    MAX = x;
                q.push(x);
            }
        }
        else if(ch == 2)
        {
            if(q.size() == 0)
                continue;
            else
                q.pop();
        }
        else
        {
            if(q.size() == 0)
                printf("0\n");
            else
                printf("%d\n",MAX);
        }
    }
    return 0;
}