二叉堆基本操作：（可用優先隊列模板）
1.上升操作（可以用於插入，並不等於插入操作）
2.下降操作（可以用於刪除，並不等於刪除操作）
3.（知道了 1和2操作）要知道怎麼刪除堆內點！
4.堆排序
二叉堆 （小堆與大堆的合併運用）典例
poj1442
Black Box
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 6131 | Accepted: 2486 |
Description
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1
N Transaction i Black Box contents after transaction Answer
(elements are arranged by non-descending)
1 ADD(3) 0 3
2 GET 1 3 3
3 ADD(1) 1 1, 3
4 GET 2 1, 3 3
5 ADD(-4) 2 -4, 1, 3
6 ADD(2) 2 -4, 1, 2, 3
7 ADD(8) 2 -4, 1, 2, 3, 8
8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8
9 GET 3 -1000, -4, 1, 2, 3, 8 1
10 GET 4 -1000, -4, 1, 2, 3, 8 2
11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.
Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.
Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.
Sample Input
7 4 3 1 -4 2 8 -1000 2 1 2 6 6
Sample Output
3 3 1 2
Source
題解：
案例解析：
輸入 M,N 分別是A數組，u數組的大小。
然後是一行A數組值，一行u數組值。
按照u數組開始，如u[1]=1 就要從A的前 u[1] 個數裏面選出第1（這個數字是u下標）小的值。->即輸出3.
同理 看u[3]=6, 就找 A前u[3]（即6）個數中 第3小的值。即輸出1；
思考：
我們這麼想：如果讓你找第n小的。那麼用小堆是不太好取到的，那麼反着想用大堆，正好大堆從底開始是 第一小的，第二小的·······到 堆頂 剛好是 第n小的。所以用大堆方便
取出所要的值。（取堆頂就OK了）。那麼大堆用永遠只有 n-1個值，然後通過小堆插入一個進來，再選出當前第n小的值。所以說每次的新插入數據，都是在選出當前最新
的第n小的值！
下面看怎麼解：
即維護 一個小堆 一個大堆；
每次用大堆來選出第n小的數，插入小堆堆頂（其實這個插入不用維護，插入的一定是小堆堆頂（即第n小的數））
然後小堆裏的 都是通過大堆選出來的 第n小的數。那麼最後輸出時，就要從這麼多第n小的數中 選出最小的。（其實就是最後一次從大堆取出來的插入小堆的數）
//體會一遍 那個用 大堆小堆的 過程吧！就會明白！
//Accepted 600K 141MS C++ 2242B
#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define MAX 30010
int M,N;
int A[MAX];
int u[MAX];
int m[MAX];//小堆
int x[MAX];//大堆
int m_num,x_num;//小堆，大堆的數據個數
void swap(int &a,int &b)
{
int t=a;a=b;b=t;
}
/**小堆**/
void adjust_min()/**下降操作（這裏只需要從頂點維護）**/
{
int lg,lr;
int x=1;
while((x<<1)<=m_num)
{
lg=x<<1;
lr=lg+1;
if(lr<=m_num&&m[lr]<m[lg])
lg=lr;
if(m[lg]<m[x])
{
swap(m[lg],m[x]);
x=lg;
}
else break;
}
}
void push_min(int x) /**上升操作（插入）**/
{
int i=++m_num;
m[m_num]=x;
while(i>1&&m[i/2]>x)
{
m[i]=m[i/2];
i/=2;
}
m[i]=x;
}
int top_min()/**得到小堆頂**/
{
return m[1];
}
void pop_min() /**刪除小堆頂**/
{
swap(m[1],m[m_num]);
m_num--;
adjust_min();
}
/**大堆(以下操作，類似於小堆.)**/
void adjust_max()
{
int lg,lr;
int key=1;
while((key<<1)<=x_num)
{
lg=key<<1;
lr=lg+1;
if(lr<=x_num&&x[lr]>x[lg])
lg=lr;
if(x[lg]>x[key])
{
swap(x[lg],x[key]);
key=lg;
}
else break;
}
}
void push_max(int key)
{
int i=++x_num;
x[x_num]=key;
while(i>1&&x[i/2]<key)
{
x[i]=x[i/2];
i/=2;
}
x[i]=key;
}
int top_max()
{
return x[1];
}
void pop_max()
{
swap(x[1],x[x_num]);
x_num--;
adjust_max();
}
/**主函數**/
int main()
{
while(~scanf("%d%d",&M,&N))
{
/**初始化堆**/
memset(m,0,sizeof(m));
memset(x,0,sizeof(x));
m_num=0;
m_num=0;
for(int i=1;i<=M;i++)
scanf("%d",&A[i]);
for(int i=1;i<=N;i++)
scanf("%d",&u[i]);
int add_index=1;
for(int i=1;i<=N;i++)
{
for(;add_index<=u[i];add_index++)
{
push_min(A[add_index]);
push_max(top_min());
pop_min();
push_min(top_max());
pop_max();
}
printf("%d\n",top_min());
push_max(top_min());
pop_min();
}
}
return 0;
}
優先隊列
#include <iostream>
#include<cstdio>
#include<queue>
using namespace std;
#define MAX 30010
int A[MAX],u[MAX];
int m,n;
int main()
{
priority_queue<int,vector<int>,greater<int> > Min;
priority_queue<int> Max;
while(~scanf("%d%d",&m,&n))
{
for(int i=1;i<=m;i++)
scanf("%d",&A[i]);
for(int i=1;i<=n;i++)
scanf("%d",&u[i]);
int add_x=1;
for(int i=1;i<=n;i++)
{
for(;add_x<=u[i];add_x++)//max 是插一次刪一次沒變，因爲它要保持堆頂爲第i小！
{ //min 是插刪插 增了一個，會選出很多個第n小到小堆，維護第n小。
Min.push(A[add_x]);
Max.push(Min.top());
Min.pop();
Min.push(Max.top());
Max.pop();
}
printf("%d\n",Min.top());
Max.push(Min.top());
Min.pop();
}
}
return 0;
}