这是我在玩弄这个数据结构和一些模板愚蠢的东西。
所有这些混乱的底部是访问两个平面数组，其中一个包含一个和树，另一个包含一个进位值树，以便以后向下传播。从概念上讲，它们形成一棵二叉树。
二叉树中节点的真实值是存储和树中的值，加上节点下的叶子数乘以从节点返回到根的所有进位树值的总和。
同时，树中每个节点的真值等于其下每个叶节点的真值。
我编写了一个函数来进行进位和求和，因为事实证明它们正在访问相同的节点。读有时会写。increase因此，您可以通过用零调用它来得到一个总和。
模板tomfoolery所做的就是计算节点在每棵树中的偏移量，以及左右子节点的位置。
虽然我确实使用了 a struct，但struct它是瞬态的——它只是一个包装器，其中包含一些预先计算的值围绕一个数组的偏移量。我确实存储了一个指向数组开头的指针，但是在这个程序中每个都block_ptr使用完全相同的值。root
对于调试，我有一些疯狂的 Assert() 和 Debug() 宏，以及递归 sum 函数调用的跟踪 nullary 函数（我用它来跟踪对其的调用总数）。再一次，为了避免全局状态而变得不必要的复杂。:)
#include <memory>
#include <iostream>
// note that you need more than 2^30 space to fit this
enum {max_tier = 30};
typedef long long intt;
#define Assert(x) (!(x)?(std::cout << "ASSERT FAILED: (" << #x << ")\n"):(void*)0)
#define DEBUG(x)
template<size_t tier, size_t count=0>
struct block_ptr
{
enum {array_size = 1+block_ptr<tier-1>::array_size * 2};
enum {range_size = block_ptr<tier-1>::range_size * 2};
intt* root;
size_t offset;
size_t logical_offset;
explicit block_ptr( intt* start, size_t index, size_t logical_loc=0 ):root(start),offset(index), logical_offset(logical_loc) {}
intt& operator()() const
{
return root[offset];
}
block_ptr<tier-1> left() const
{
return block_ptr<tier-1>(root, offset+1, logical_offset);
}
block_ptr<tier-1> right() const
{
return block_ptr<tier-1>(root, offset+1+block_ptr<tier-1>::array_size, logical_offset+block_ptr<tier-1>::range_size);
}
enum {is_leaf=false};
};
template<>
struct block_ptr<0>
{
enum {array_size = 1};
enum {range_size = 1};
enum {is_leaf=true};
intt* root;
size_t offset;
size_t logical_offset;
explicit block_ptr( intt* start, size_t index, size_t logical_loc=0 ):root(start),offset(index), logical_offset(logical_loc)
{}
intt& operator()() const
{
return root[offset];
}
// exists only to make some of the below code easier:
block_ptr<0> left() const { Assert(false); return *this; }
block_ptr<0> right() const { Assert(false); return *this; }
};
template<size_t tier>
void propogate_carry( block_ptr<tier> values, block_ptr<tier> carry )
{
if (carry() != 0)
{
values() += carry() * block_ptr<tier>::range_size;
if (!block_ptr<tier>::is_leaf)
{
carry.left()() += carry();
carry.right()() += carry();
}
carry() = 0;
}
}
// sums the values from begin to end, but not including end!
// ie, the half-open interval [begin, end) in the tree
// if increase is non-zero, increases those values by that much
// before returning it
template<size_t tier, typename trace>
intt query_or_modify( block_ptr<tier> values, block_ptr<tier> carry, int begin, int end, int increase=0, trace const& tr = [](){} )
{
tr();
DEBUG(
std::cout << begin << " " << end << " " << increase << "\n";
if (increase)
{
std::cout << "Increasing " << end-begin << " elements by " << increase << " starting at " << begin+values.offset << "\n";
}
else
{
std::cout << "Totaling " << end-begin << " elements starting at " << begin+values.logical_offset << "\n";
}
)
if (end <= begin)
return 0;
size_t mid = block_ptr<tier>::range_size / 2;
DEBUG( std::cout << "[" << values.logical_offset << ";" << values.logical_offset+mid << ";" << values.logical_offset+block_ptr<tier>::range_size << "]\n"; )
// exatch math first:
bool bExact = (begin == 0 && end >= block_ptr<tier>::range_size);
if (block_ptr<tier>::is_leaf)
{
Assert(bExact);
}
bExact = bExact || block_ptr<tier>::is_leaf; // leaves are always exact
if (bExact)
{
carry()+=increase;
intt retval = (values()+carry()*block_ptr<tier>::range_size);
DEBUG( std::cout << "Exact sum is " << retval << "\n"; )
return retval;
}
// we don't have an exact match. Apply the carry and pass it down to children:
propogate_carry(values, carry);
values() += increase * end-begin;
// Now delegate to children:
if (begin >= mid)
{
DEBUG( std::cout << "Right:"; )
intt retval = query_or_modify( values.right(), carry.right(), begin-mid, end-mid, increase, tr );
DEBUG( std::cout << "Right sum is " << retval << "\n"; )
return retval;
}
else if (end <= mid)
{
DEBUG( std::cout << "Left:"; )
intt retval = query_or_modify( values.left(), carry.left(), begin, end, increase, tr );
DEBUG( std::cout << "Left sum is " << retval << "\n"; )
return retval;
}
else
{
DEBUG( std::cout << "Left:"; )
intt left = query_or_modify( values.left(), carry.left(), begin, mid, increase, tr );
DEBUG( std::cout << "Right:"; )
intt right = query_or_modify( values.right(), carry.right(), 0, end-mid, increase, tr );
DEBUG( std::cout << "Right sum is " << left << " and left sum is " << right << "\n"; )
return left+right;
}
}
这里有一些帮助类可以轻松创建给定大小的段树。但是请注意，您所需要的只是一个大小合适的数组，并且可以从指向元素 0 的指针构造一个 block_ptr，然后就可以开始了。
template<size_t tier>
struct segment_tree
{
typedef block_ptr<tier> full_block_ptr;
intt block[full_block_ptr::range_size];
full_block_ptr root() { return full_block_ptr(&block[0],0); }
void init()
{
std::fill_n( &block[0], size_t(full_block_ptr::range_size), 0 );
}
};
template<size_t entries, size_t starting=0>
struct required_tier
{
enum{ tier =
block_ptr<starting>::array_size >= entries
?starting
:required_tier<entries, starting+1>::tier
};
enum{ error =
block_ptr<starting>::array_size >= entries
?false
:required_tier<entries, starting+1>::error
};
};
// max 2^30, to limit template generation.
template<size_t entries>
struct required_tier<entries, size_t(max_tier)>
{
enum{ tier = 0 };
enum{ error = true };
};
// really, these just exist to create an array of the correct size
typedef required_tier< 1000000 > how_big;
enum {tier = how_big::tier};
int main()
{
segment_tree<tier> values;
segment_tree<tier> increments;
Assert(!how_big::error); // can be a static assert -- fails if the enum of max tier is too small for the number of entries you want
values.init();
increments.init();
auto value_root = values.root();
auto carry_root = increments.root();
size_t count = 0;
auto tracer = [&count](){count++;};
intt zero = query_or_modify( value_root, carry_root, 0, 100000, 0, tracer );
std::cout << "zero is " << zero << " in " << count << " steps\n";
count = 0;
Assert( zero == 0 );
intt test2 = query_or_modify( value_root, carry_root, 0, 100, 10, tracer ); // increase everything from 0 to 100 by 10
Assert(test2 == 1000);
std::cout << "test2 is " << test2 << " in " << count << " steps \n";
count = 0;
intt test3 = query_or_modify( value_root, carry_root, 1, 1000, 0, tracer );
Assert(test3 == 990);
std::cout << "test3 is " << test3 << " in " << count << " steps\n";
count = 0;
intt test4 = query_or_modify( value_root, carry_root, 50, 5000, 87, tracer );
Assert(test4 == 10*(100-50) + 87*(5000-50) );
std::cout << "test4 is " << test4 << " in " << count << " steps\n";
count = 0;
}
虽然这不是您想要的答案，但它可能会使某人更容易编写它。写这个让我很开心。所以，希望对您有所帮助！
该代码在 Ideone.com 上使用 C++0x 编译器进行了测试和编译。