hdu 5349 的传送门

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

题目大意：给你一个数，表示有几行，然后给你一个a和x，如果a等于1的话，就向里面增加一个数，如果a等于2的话 ，就删除集合中最小的一个数，如果等于a==3 的话，就输出最大的数；
解题思路：STL,集合；

具体见代码：

#include <iostream>
#include <cstdio>
#include <set>
using namespace std;
set<long long >::iterator it;
set<long long > s;

int main()
{
    long long n, a, x;
    scanf("%lld",&n);
    s.clear();
    while(n--)
    {
        scanf("%lld",&a);
        if(a == 1)
        {
            scanf("%lld",&x);
            s.insert(x);
        }
        else if(a == 2)
        {
            if(!s.empty())
               s.erase(s.begin());
        }
        else if(a == 3)
        {
            if(s.empty())
                puts("0");
            else
            {
                it=s.end();
                it--;
                printf("%lld\n",*it);
            }
        }
    }
    return 0;
}