Online minimization problem

Description

To a space-separated string, the string only four characters, respectively:
(1) space, representing the separator;
(2) an integer value representing the currently inserted;
(3) the letter E, represents from previously removal of a minimum digital value;
(4) the decimal point indicates the end of input. "";
known insert has n, m one operand removed, find the number of m sequentially acquired, taken to ensure that each input It has a number of desirable hours. (Between 1≤m≤n≤100000 each number in the range -10000 and 10000)

Input

Only his string

Output

Only one row of data, separated by spaces of m numbers, respectively, the minimum value of m sequentially obtained.

Sample Input

4 8 E 3 E 10 2 6 E 1.

Sample Output

4 3 2

HINT

n <= 200, all the input data are <= 1000, is less than the determined minimum of 10 ^ 9.

Source

#include<bits/stdc++.h>
using namespace std;
int s[100001],len;
void s_up(int p)
{
    while(p>1&&s[p/2]>s[p])
    {
        swap(s[p/2],s[p]);
        p=p/2;
    }
    return;
}
void s_down(int p)
{
    int lt;
    while(1)
    {
        if(p*2>len) return;
        if(p*2==len) lt=p*2;
        else
        {
            if(s[p*2]<s[p*2+1]) 
            lt=p*2;
            else
            lt=p*2+1;
        }
        if(s[p]>s[lt])
        {
            swap(s[p],s[lt]);
            p=lt;
        }
        else break;
    }
    return;
}
void insert(int key)
{
    len++;
    s[len]=key;
    s_up(len);
}
int main()
{
    char c;
    int num=0;
    bool flag=0;
    int fh=1;
    while(1){
        c=getchar();
        if(c=='.') break;
        else if(c=='E') 
        {
            printf("%d ",s[1]);
            s[1]=s[len--];
            s_down(1);
        }
        else if(c==' ')
        {
            if(flag==1)
            {
                insert(fh*num);
                num=0;
                flag=0;
                fh=1;
            }
        }
        else
        if(c=='-') fh=-1;
        else{
            flag=1;
            num=num*10+c-'0';
            }
    }
    return 0;
}