The kth great number

最新推荐文章于 2019-03-13 23:43:34 发布

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最新推荐文章于 2019-03-13 23:43:34 发布

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原文链接：http://www.cnblogs.com/liyakun/archive/2013/04/20/3033141.html

小明和小宝正在玩一个数字游戏。小明可以选择写下数字或询问小宝第k大的数字。小宝感到困惑，尝试帮助他解决游戏中的数字排序问题。

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Problem Description

Xiao Ming and Xiao Bao are playing a simple Numbers game. In a round Xiao Ming can choose to write down a number, or ask Xiao Bao what the kth great number is. Because the number written by Xiao Ming is too much, Xiao Bao is feeling giddy. Now, try to help Xiao Bao.

Input

There are several test cases. For each test case, the first line of input contains two positive integer n, k. Then n lines follow. If Xiao Ming choose to write down a number, there will be an " I" followed by a number that Xiao Ming will write down. If Xiao Ming choose to ask Xiao Bao, there will be a "Q", then you need to output the kth great number.

Output

The output consists of one integer representing the largest number of islands that all lie on one line.

#include<iostream>
#include<cstring>
using namespace std;
int heap[10000000];
void swap(int &a,int &b);
void put(int &len,int m);
void delet(int &len);
int main()
{
    int n,k,m,i,j,len=0;
    char com;
    while(cin>>n>>k)
    {
        len=0;
        while(n--)
        {
            cin>>com;
            if(com=='I')
            {
                cin>>m;
                if(len<k)
                put(len,m);
                else
                {
                    if(heap[1]<m)
                    {
                        delet(len);
                        put(len,m);
                    }
                }
            }
            else
            {
                cout<<heap[1]<<endl;
            }
        }
    }
}
void swap(int &a,int &b)
{
    int temp;
    temp=a;
    a=b;
    b=temp;
}
void put(int &len,int m)
{
    len++;
    heap[len]=m;
    int son=len;
    while(son!=1&&heap[son]<heap[son/2])
    {
        swap(heap[son],heap[son/2]);
        son/=2;
    }
}
void delet(int &len)
{
    heap[1]=heap[len];
    --len;
    int son=1;
    while(son*2<=len)
    {
        if(son*2+1>len||heap[son*2]<heap[son*2+1])
        son=son*2;
        else
        son=son*2+1;
        if(heap[son/2]>heap[son])
        swap(heap[son],heap[son/2]);
        else
        break;
    }
}

转载于:https://www.cnblogs.com/liyakun/archive/2013/04/20/3033141.html