SET的第一行模板原始碼是這樣的：

  template<typename _Key, typename _Compare = std::less<_Key>,
	   typename _Alloc = std::allocator<_Key> >

根據template和English的有關知識，我們可以知道set的定義方式應該是：
set<型別，比較函式（不寫時預設為less<型別>，從小到大），空間配置器（從應用STL角度來說完全可以忽略，實際上也完全一定要忽略不寫）>變數名

第110行：

      typedef _Rb_tree<key_type, value_type, _Identity<value_type>,
		       key_compare, _Key_alloc_type> _Rep_type;
      _Rep_type _M_t;  // Red-black tree representing set.
 

現在我們知道set是用紅黑樹來維護的。

按順序 第149行：

      /**
       *  @brief  Builds a %set from a range.
       *  @param  first  An input iterator.
       *  @param  last  An input iterator.
       *
       *  Create a %set consisting of copies of the elements from [first,last).
       *  This is linear in N if the range is already sorted, and NlogN
       *  otherwise (where N is distance(first,last)).
       */
      template<typename _InputIterator>
        set(_InputIterator __first, _InputIterator __last)
	: _M_t()
        { _M_t._M_insert_unique(__first, __last); }

      /**
       *  @brief  Builds a %set from a range.
       *  @param  first  An input iterator.
       *  @param  last  An input iterator.
       *  @param  comp  A comparison functor.
       *  @param  a  An allocator object.
       *
       *  Create a %set consisting of copies of the elements from [first,last).
       *  This is linear in N if the range is already sorted, and NlogN
       *  otherwise (where N is distance(first,last)).
       */
      template<typename _InputIterator>
        set(_InputIterator __first, _InputIterator __last,
	    const _Compare& __comp,
	    const allocator_type& __a = allocator_type())
	: _M_t(__comp, __a)
        { _M_t._M_insert_unique(__first, __last); }
 

這個的意思是可以用迭代器初始化set（不懂可以看下面的測試程式碼），並且速度在資料有序的情況下為O(n)(常數偏大),無序的情況下為O(nlogn)（常數偏小）。

我的測試程式碼：

#include<bits/stdc++.h>
using namespace std;

vector<int>vec;
set<int>st;
int main()
{
	for(int i=1;i<=2000000;i++)
		vec.push_back(i);
	int t1 = clock();
	st = set<int>(vec.begin(),vec.end());
	printf("%d\n",clock()-t1);
	
	random_shuffle(vec.begin(),vec.end());
	t1=clock();
	st = set<int>(vec.begin(),vec.end());
	printf("%d\n",clock()-t1);
}
 

輸出（不開O2）：
1083
3007
輸出（開O2）：
724
1689
可以發現在開O2的情況下還是可以接受的，
100000個數只需要100ms左右。

219行

      /**
       *  @brief  %Set assignment operator.
       *  @param  x  A %set of identical element and allocator types.
       *
       *  All the elements of @a x are copied, but unlike the copy constructor,
       *  the allocator object is not copied.
       */
      set&
      operator=(const set& __x)
      {
	_M_t = __x._M_t;
	return *this;
      }

set的等於符號過載，O(n)
測試程式碼：

#include<bits/stdc++.h>
using namespace std;

vector<int>vec;
set<int>st;
int main()
{
	for(int i=1;i<=100000;i++)
		vec.push_back(i);
	int t1 = clock();
	st = set<int>(vec.begin(),vec.end());
	printf("%d\n",clock()-t1);
	
	t1=clock();
	set<int>st2 = st;
	printf("%d\n",clock()-t1);
}

輸出 ：
57
18

689行：

  template<typename _Key, typename _Compare, typename _Alloc>
    inline void
    swap(set<_Key, _Compare, _Alloc>& __x, set<_Key, _Compare, _Alloc>& __y)
    { __x.swap(__y); }

這個swap好像是直接交換指標的，O(1)

測試程式碼：

#include<bits/stdc++.h>
using namespace std;

vector<int>vec;
set<int>st;
int main()
{
	for(int i=1;i<=100000;i++)
		vec.push_back(i);
	int t1 = clock();
	st = set<int>(vec.begin(),vec.end());
	printf("%d\n",clock()-t1);
	
	t1=clock();
	set<int>st2;
	st2.swap(st);
	printf("%d\n",*st2.begin());
	printf("%d\n",clock()-t1);
}

輸出：
57
1
0

943行：

      /**
       *  Erases all the elements.  Note that this function only erases the
       *  elements, and that if the elements themselves are pointers, the
       *  pointed-to memory is not touched in any way.  Managing the pointer is
       *  the user's responsibility.
       */
      void
      clear()
      { _M_erase_at_end(this->_M_impl._M_start); }

clear()函式，O(n)的。（Managing the pointer is the user’s responsibility.這話好絕，再也不用set裝指標了）

測試程式碼：

#include<bits/stdc++.h>
using namespace std;

vector<int>vec;
set<int>st;
int main()
{
	for(int i=1;i<=100000;i++)
		vec.push_back(i);
	int t1 = clock();
	st = set<int>(vec.begin(),vec.end());
	printf("%d\n",clock()-t1);
	
	t1=clock();
	st.clear();
	printf("%d\n",clock()-t1);
}

輸出：
53
10

272行：

      ///  Returns the comparison object with which the %set was constructed.
      key_compare
      key_comp() const
      { return _M_t.key_comp(); }

現在來解釋一下比較函式。
因為STL比較強大，比較函式並不是一個函式，而是一個結構體，裡面過載了（）符號，用（a,b）代表比較a和b，並且返回一個值，1代表a<b，0otherwise（被STL註釋語言同化了）
具體看下面程式碼應該能夠有一些瞭解，需要一些template和結構體的前置知識。

測試程式碼：

#include<bits/stdc++.h>
using namespace std;

vector<int>vec;
set<int>st;

struct cmp
{
	bool operator ()(int u,int v){ return u<v; }
}a;

int main()
{
	printf("%d\n",a(2,1));
	printf("%d\n",a(1,2));
	less<int>b;
	printf("%d\n",b(1,2));
	greater<int>c;
	printf("%d\n",c(1,2));
}

輸出：
0
1
1
0

285行，8種迭代器：

      /**
       *  Returns a read-only (constant) iterator that points to the first
       *  element in the %set.  Iteration is done in ascending order according
       *  to the keys.
       */
      iterator
      begin() const
      { return _M_t.begin(); }

      /**
       *  Returns a read-only (constant) iterator that points one past the last
       *  element in the %set.  Iteration is done in ascending order according
       *  to the keys.
       */
      iterator
      end() const
      { return _M_t.end(); }

      /**
       *  Returns a read-only (constant) iterator that points to the last
       *  element in the %set.  Iteration is done in descending order according
       *  to the keys.
       */
      reverse_iterator
      rbegin() const
      { return _M_t.rbegin(); }

      /**
       *  Returns a read-only (constant) reverse iterator that points to the
       *  last pair in the %set.  Iteration is done in descending order
       *  according to the keys.
       */
      reverse_iterator
      rend() const
      { return _M_t.rend(); }

#ifdef __GXX_EXPERIMENTAL_CXX0X__
      /**
       *  Returns a read-only (constant) iterator that points to the first
       *  element in the %set.  Iteration is done in ascending order according
       *  to the keys.
       */
      iterator
      cbegin() const
      { return _M_t.begin(); }

      /**
       *  Returns a read-only (constant) iterator that points one past the last
       *  element in the %set.  Iteration is done in ascending order according
       *  to the keys.
       */
      iterator
      cend() const
      { return _M_t.end(); }

      /**
       *  Returns a read-only (constant) iterator that points to the last
       *  element in the %set.  Iteration is done in descending order according
       *  to the keys.
       */
      reverse_iterator
      crbegin() const
      { return _M_t.rbegin(); }

      /**
       *  Returns a read-only (constant) reverse iterator that points to the
       *  last pair in the %set.  Iteration is done in descending order
       *  according to the keys.
       */
      reverse_iterator
      crend() const
      { return _M_t.rend(); }

好像前面加不加c都一樣。。。迭代器就是一個變數只支援++和–，即只能迭代到下一個或上一個，因為這像迭代的過程，故名迭代器。
begin()是第一個，end()是最後一個的後一個（即是空），體現了STL左閉右開的風格。
rbegin()是最後一個，rend()是最前面的前面一個（即是空），體現了STL左閉右開的風格。

359行：

      ///  Returns true if the %set is empty.
      bool
      empty() const
      { return _M_t.empty(); }

      ///  Returns the size of the %set.
      size_type
      size() const
      { return _M_t.size(); }

size和empty，都是O(1)的。

#include<bits/stdc++.h>
using namespace std;

vector<int>vec;
set<int>st;
int main()
{
	for(int i=1;i<=1000000;i++) st.insert(i);
	int t1 = clock();
	printf("%d\n",st.size());
	printf("%d\n",st.empty());
	printf("%d\n",clock()-t1);
}

輸出：
1000000
0
0
393行

      // insert/erase
      /**
       *  @brief Attempts to insert an element into the %set.
       *  @param  x  Element to be inserted.
       *  @return  A pair, of which the first element is an iterator that points
       *           to the possibly inserted element, and the second is a bool
       *           that is true if the element was actually inserted.
       *
       *  This function attempts to insert an element into the %set.  A %set
       *  relies on unique keys and thus an element is only inserted if it is
       *  not already present in the %set.
       *
       *  Insertion requires logarithmic time.
       */
      std::pair<iterator, bool>
      insert(const value_type& __x)
      {
	std::pair<typename _Rep_type::iterator, bool> __p =
	  _M_t._M_insert_unique(__x);
	return std::pair<iterator, bool>(__p.first, __p.second);
      }

insert是有返回值的！！！Insertion requires logarithmic time。。。
看註釋應該就懂了，相同元素不重複插入，pair<set::iterater , bool>的前一個是插入的元素的迭代器，後一個是插入是否成功。
測試程式碼

#include<bits/stdc++.h>
using namespace std;

vector<int>vec;
set<int>st;
int main()
{
	pair<set<int>::iterator , bool>tmp = st.insert(1);
	printf("%d %d\n",*tmp.first,tmp.second);
	pair<set<int>::iterator , bool>res = st.insert(1);
	printf("%d %d\n",*res.first,res.second);
	pair<set<int>::iterator , bool>res1 = st.insert(2);
	printf("%d %d\n",*res1.first,res1.second);
}

輸出：
1 1
1 0
2 1

415行：

      /**
       *  @brief Attempts to insert an element into the %set.
       *  @param  position  An iterator that serves as a hint as to where the
       *                    element should be inserted.
       *  @param  x  Element to be inserted.
       *  @return  An iterator that points to the element with key of @a x (may
       *           or may not be the element passed in).
       *
       *  This function is not concerned about whether the insertion took place,
       *  and thus does not return a boolean like the single-argument insert()
       *  does.  Note that the first parameter is only a hint and can
       *  potentially improve the performance of the insertion process.  A bad
       *  hint would cause no gains in efficiency.
       *
       *  For more on "hinting", see:
       *  http://gcc.gnu.org/onlinedocs/libstdc++/manual/bk01pt07ch17.html
       *  
       *  Insertion requires logarithmic time (if the hint is not taken).
       */
      iterator
      insert(iterator __position, const value_type& __x)
      { return _M_t._M_insert_unique_(__position, __x); }

指定位置的insert，能夠potentially improve the performance of the insertion process.（卡常必備？？？）
測試程式碼：

#include<bits/stdc++.h>
using namespace std;

vector<int>vec;
set<int>st;
set<int>::iterator it;
int main()
{
	st.insert(1);
	int t1=clock();
	it = st.begin();
	for(int i=2;i<=100000;i++)
	{
		
		it++;
		st.insert(it,i);
		
		/*
		st.insert(i);
		*/
	}
	printf("%d\n",clock()-t1);
	printf("%d\n",st.size());

	st.clear();
	st.insert(1);
	t1=clock();
	it = st.begin();
	for(int i=2;i<=100000;i++)
	{
		/*
		it++;
		st.insert(it,i);
		*/
		
		st.insert(i);
		
	}
	printf("%d\n",clock()-t1);
	printf("%d\n",st.size());
}

輸出：
30
100000
74
100000
效果好像還不錯，vector初始化好像要55ms來著。。。

465行：

      /**
       *  @brief Erases an element from a %set.
       *  @param  position  An iterator pointing to the element to be erased.
       *
       *  This function erases an element, pointed to by the given iterator,
       *  from a %set.  Note that this function only erases the element, and
       *  that if the element is itself a pointer, the pointed-to memory is not
       *  touched in any way.  Managing the pointer is the user's responsibility.
       */
      void
      erase(iterator __position)
      { _M_t.erase(__position); }

      /**
       *  @brief Erases elements according to the provided key.
       *  @param  x  Key of element to be erased.
       *  @return  The number of elements erased.
       *
       *  This function erases all the elements located by the given key from
       *  a %set.
       *  Note that this function only erases the element, and that if
       *  the element is itself a pointer, the pointed-to memory is not touched
       *  in any way.  Managing the pointer is the user's responsibility.
       */
      size_type
      erase(const key_type& __x)
      { return _M_t.erase(__x); }

      /**
       *  @brief Erases a [first,last) range of elements from a %set.
       *  @param  first  Iterator pointing to the start of the range to be
       *                 erased.
       *  @param  last  Iterator pointing to the end of the range to be erased.
       *
       *  This function erases a sequence of elements from a %set.
       *  Note that this function only erases the element, and that if
       *  the element is itself a pointer, the pointed-to memory is not touched
       *  in any way.  Managing the pointer is the user's responsibility.
       */
      void
      erase(iterator __first, iterator __last)
      { _M_t.erase(__first, __last); }

用迭代器無返回值，用值刪除會返回刪除的個數（意味著用multiset的會將同一個值的刪完），迭代器區間刪除，作者也懶得分析複雜度了。

測試程式碼：

#include<bits/stdc++.h>
using namespace std;

vector<int>vec;
set<int>st;
set<int>::iterator it;
int main()
{
	st.insert(1);
	it = st.begin();
	for(int i=2;i<=1000000;i++)
	{
		it++;
		st.insert(it,i);
	}
	int t1=clock();
	printf("%d\n",st.erase(233));
	printf("%d\n",st.erase(233));
	for(int i=1;i<=100000;i++)
		st.erase(i);
	printf("%d\n",clock()-t1);
}

輸出：
1
0
67
刪除常數也可以接受吧。
518行：

      // set operations:

      /**
       *  @brief  Finds the number of elements.
       *  @param  x  Element to located.
       *  @return  Number of elements with specified key.
       *
       *  This function only makes sense for multisets; for set the result will
       *  either be 0 (not present) or 1 (present).
       */
      size_type
      count(const key_type& __x) const
      { return _M_t.find(__x) == _M_t.end() ? 0 : 1; }

這程式碼好露骨啊。
535行：

      /**
       *  @brief Tries to locate an element in a %set.
       *  @param  x  Element to be located.
       *  @return  Iterator pointing to sought-after element, or end() if not
       *           found.
       *
       *  This function takes a key and tries to locate the element with which
       *  the key matches.  If successful the function returns an iterator
       *  pointing to the sought after element.  If unsuccessful it returns the
       *  past-the-end ( @c end() ) iterator.
       */
      iterator
      find(const key_type& __x)
      { return _M_t.find(__x); }

      const_iterator
      find(const key_type& __x) const
      { return _M_t.find(__x); }

find函式，返回迭代器，沒有就返回end()；
關於後面那個函式我的理解是，set如果是const的就返回const_iterator,歡迎評論。

557行：

      /**
       *  @brief Finds the beginning of a subsequence matching given key.
       *  @param  x  Key to be located.
       *  @return  Iterator pointing to first element equal to or greater
       *           than key, or end().
       *
       *  This function returns the first element of a subsequence of elements
       *  that matches the given key.  If unsuccessful it returns an iterator
       *  pointing to the first element that has a greater value than given key
       *  or end() if no such element exists.
       */
      iterator
      lower_bound(const key_type& __x)
      { return _M_t.lower_bound(__x); }

      const_iterator
      lower_bound(const key_type& __x) const
      { return _M_t.lower_bound(__x); }
      //@}

      //@{
      /**
       *  @brief Finds the end of a subsequence matching given key.
       *  @param  x  Key to be located.
       *  @return Iterator pointing to the first element
       *          greater than key, or end().
       */
      iterator
      upper_bound(const key_type& __x)
      { return _M_t.upper_bound(__x); }

      const_iterator
      upper_bound(const key_type& __x) const
      { return _M_t.upper_bound(__x); }

lower_bound()和upper_bound()，意義顯然。

594行：

       *  @brief Finds a subsequence matching given key.
       *  @param  x  Key to be located.
       *  @return  Pair of iterators that possibly points to the subsequence
       *           matching given key.
       *
       *  This function is equivalent to
       *  @code
       *    std::make_pair(c.lower_bound(val),
       *                   c.upper_bound(val))
       *  @endcode
       *  (but is faster than making the calls separately).
       *
       *  This function probably only makes sense for multisets.
       */
      std::pair<iterator, iterator>
      equal_range(const key_type& __x)
      { return _M_t.equal_range(__x); }

      std::pair<const_iterator, const_iterator>
      equal_range(const key_type& __x) const
      { return _M_t.equal_range(__x); }

一次計算出lower_bound(x)和upper_bound(x)況且只用一次的時間。。。。好像很有用（嗎？）。equal_range…

627行:

  /**
   *  @brief  Set equality comparison.
   *  @param  x  A %set.
   *  @param  y  A %set of the same type as @a x.
   *  @return  True iff the size and elements of the sets are equal.
   *
   *  This is an equivalence relation.  It is linear in the size of the sets.
   *  Sets are considered equivalent if their sizes are equal, and if
   *  corresponding elements compare equal.
  */
  template<typename _Key, typename _Compare, typename _Alloc>
    inline bool
    operator==(const set<_Key, _Compare, _Alloc>& __x,
	       const set<_Key, _Compare, _Alloc>& __y)
    { return __x._M_t == __y._M_t; }

  /**
   *  @brief  Set ordering relation.
   *  @param  x  A %set.
   *  @param  y  A %set of the same type as @a x.
   *  @return  True iff @a x is lexicographically less than @a y.
   *
   *  This is a total ordering relation.  It is linear in the size of the
   *  maps.  The elements must be comparable with @c <.
   *
   *  See std::lexicographical_compare() for how the determination is made.
  */
  template<typename _Key, typename _Compare, typename _Alloc>
    inline bool
    operator<(const set<_Key, _Compare, _Alloc>& __x,
	      const set<_Key, _Compare, _Alloc>& __y)
    { return __x._M_t < __y._M_t; }

set的大小比較，<號被定義為字典序？？？線性複雜度。
multiset就是支援重複元素而已，記得erase的區別，很致命。。。。
完