2015 Multi-University Training Contest 5 1007

MZL's simple problem

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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

 

题意：三种对集合的操作：1、加入一个数 2、删除最小值 3、求最大值

分析：加入时，维护最大值，删除时如果集合为空，则定义最大值为无穷小

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<string>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<stdlib.h>
#include<algorithm>
#define LL __int64
using namespace std;
const int INF=0x3f3f3f3f;
int n;
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        int opt,x;
        int maxn=-INF;
        int f=0,r=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&opt);
            if(opt==1)
            {
                scanf("%d",&x);
                r++;
                if(x>maxn) maxn=x;
            }
            if(opt==2)
            {
                if(f<r) f++;
                if(f==r) maxn=-INF;
            }
            if(opt==3)
            {
                if(f==r) printf("0\n");
                else printf("%d\n",maxn);
            }
        }
    }
    return 0;
}