HDU 5349 MZL's simple problem（优先队列）

MZL's simple problem

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 776 Accepted Submission(s): 375



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

Source

2015 Multi-University Training Contest 5



点击打开链接

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<string>
#include<math.h>
#include<queue>
#include<stack>
#include<map>
#define INF 0x3f3f3f3f
#define eps 1e-6

using namespace std;

priority_queue<int, vector<int>, greater<int> > q;///从小到大排列

int n;
int main()
{
    while(scanf("%d",&n)!=EOF)
    {

        while(!q.empty())
        {
            q.pop();
        }
        int x,y;
        int flag  = 0;
        int maxx;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&x);
            if(x == 1)
            {
                scanf("%d",&y);
                if(flag == 0)
                {
                    flag = 1;
                    maxx = y;
                }
                if(maxx < y)
                {
                    maxx = y;
                }
                q.push(y);
            }
            else if(x == 2)
            {
                if(q.size() > 0)
                {
                    int tt = q.top();
                    while(!q.empty() && q.top() == tt)
                    {
                        q.pop();
                    }
                }
                if(q.size() <= 0)
                {
                    flag = 0;
                }
            }
            else
            {
                if(q.size() > 0)
                {
                    printf("%d\n",maxx);
                }
                else
                {
                    printf("0\n");
                }
            }
        }
    }
    return 0;
}