堆的实现 2196. Nuanran's Idol II

堆用数组实现，时间复杂度为O(logn),如果进一本书就排序一次的话时间复杂度很高。

You have known that nuanran is a loyal fan of Kelly from the last contest. For this reason, nuanran is interested in collecting pictures of Kelly. Of course, he doesn't like each picture equally. So he gives each picture a score which is a positive integer. And the larger the score is, the more he likes that picture. Nuanran often goes to buy new pictures in some shops to increase his picture collection.

Sometimes nuanran's friends ask him for Kelly's picture and he will choose the picture having the smallest score. Because he has many pictures, this is a boring task. Can you help him again?

Input

For each test case, the first line gives an integer n (1 ≤ n ≤ 100000). Then n lines follow, each line has one of the following two formats.

"B S". Denoting nuanran buys a new picture and S is the score of that picture.

"G". Denoting nuanran gives a picture to his friends.

The input is terminated when n=0.

Output

For each giving test case, you should output the score of the picture nuanran gives to his friends.

Sample Input

8B 20B 10GB 9GB 100B 25G0

Sample Output

10920

#include<iostream>#include<cstring>using namespace std;int a[100001],len=0;//len为堆的尾部int leftchild(int n){return n*2;}int rightchild(int n){return n*2+1;}int parent(int n){return n/2;}int gettop(){return a[1];}void heapfy(int t){int l,r,m;l=leftchild(t);r=rightchild(t);m=t;if(l<len&&a[m]>a[l])m=l;if(r<len&&a[m]>a[r])m=r;//寻找最小值下标if(m!=t){swap(a[t],a[m]);heapfy(m);}return;}void insert(int t){a[len]=t;int p=parent(len);int q=len;while(a[p]>a[q]&&p>0){swap(a[p],a[q]);q=p;p=parent(p);}len++;return ;}void pop(){if(len==0)return;len--;//cout<<a[1];a[1]=a[len];a[len]=0;heapfy(1);return;}int main(){int n,s;char c;while(cin>>n&&n){len=1;memset(a,-1,sizeof(a));while(n--){cin>>c;if(c=='B'){cin>>s;insert(s);}else{cout<<gettop()<<endl;pop();}}}return 0;}