查找水平线段和垂直线段之间的三角形数量
先决条件： BIT给定'n'条线段，每条线段要么是水平的，要么是垂直的，求这些线段的交点连接起来可以形成的三角形（包括面积为零的三角形）的最大数目。
没有两条水平线段重叠，也没有两条垂直线段重叠。一条线用两个点表示（四个整数，前两个分别是第一个点的 x 和 y 坐标，另外两个分别是第二个点的 x 和 y 坐标）
例子：
|
---|-------|--
| | -----
| --|--|- |
| | |
For the above line segments, there are four points of intersection between
vertical and horizontal lines, every three out of which form a triangle,
so there can be 4C3 triangles.
该想法基于扫描线算法。
分步构建解决方案：
- 将具有相应事件（如下所述）的所有线段的两个点存储在一个向量中，并按其 x 坐标的非递减顺序对所有点进行排序。
- 现在让我们想象一条垂直线，我们扫过所有这些点并根据我们当前所在的点描述 3 个事件：
- in – 水平线段的最左边点
- out – 水平线段的最右边点
- 一条垂直线
- 我们称之为有过的第一个事件，但没有第二区“主动”或横线“活动”。我们将有一个 BIT（二进制索引树）来存储所有活动线的“y”坐标。
- 一旦一行变为非活动状态，我们就从 BIT 中删除它的“y”。
- 当第三类事件发生时，即当我们在一条垂直线上时，我们在它的 'y' 坐标范围内查询树，并将结果与到目前为止的交点数相加。
- 最后，我们将有交点的数量，比如m ，那么三角形（包括零面积）的数量将是m C 3 。
注意：我们需要仔细排序点，看实现中的cmp()函数进行澄清。
// A C++ implementation of the above idea
#include
#define maxy 1000005
#define maxn 10005
using namespace std;
// structure to store point
struct point
{
int x, y;
point(int a, int b)
{
x = a, y = b;
}
};
// Note: Global arrays are initially zero
// array to store BIT and vector to store
// the points and their corresponding event number,
// in the second field of the pair
int bit[maxy];
vector > events;
// compare function to sort in order of non-decreasing
// x coordinate and if x coordinates are same then
// order on the basis of events on the points
bool cmp(pair &a, pair &b)
{
if ( a.first.x != b.first.x )
return a.first.x < b.first.x;
//if the x coordinates are same
else
{
// both points are of the same vertical line
if (a.second == 3 && b.second == 3)
{
return true;
}
// if an 'in' event occurs before 'vertical'
// line event for the same x coordinate
else if (a.second == 1 && b.second == 3)
{
return true;
}
// if a 'vertical' line comes before an 'in'
// event for the same x coordinate, swap them
else if (a.second == 3 && b.second == 1)
{
return false;
}
// if an 'out' event occurs before a 'vertical'
// line event for the same x coordinate, swap.
else if (a.second == 2 && b.second == 3)
{
return false;
}
//in all other situations
return true;
}
}
// update(y, 1) inserts a horizontal line at y coordinate
// in an active region, while update(y, -1) removes it
void update(int idx, int val)
{
while (idx < maxn)
{
bit[idx] += val;
idx += idx & (-idx);
}
}
// returns the number of lines in active region whose y
// coordinate is between 1 and idx
int query(int idx)
{
int res = 0;
while (idx > 0)
{
res += bit[idx];
idx -= idx & (-idx);
}
return res;
}
// inserts a line segment
void insertLine(point a, point b)
{
// if it is a horizontal line
if (a.y == b.y)
{
int beg = min(a.x, b.x);
int end = max(a.x, b.x);
// the second field in the pair is the event number
events.push_back(make_pair(point(beg, a.y), 1));
events.push_back(make_pair(point(end, a.y), 2));
}
//if it is a vertical line
else
{
int up = max(b.y, a.y);
int low = min(b.y, a.y);
//the second field of the pair is the event number
events.push_back(make_pair(point(a.x, up), 3));
events.push_back(make_pair(point(a.x, low), 3));
}
}
// returns the number of intersection points between all
// the lines, vertical and horizontal, to be run after the
// points have been sorted using the cmp() function
int findIntersectionPoints()
{
int intersection_pts = 0;
for (int i = 0 ; i < events.size() ; i++)
{
//if the current point is on an 'in' event
if (events[i].second == 1)
{
//insert the 'y' coordinate in the active region
update(events[i].first.y, 1);
}
// if current point is on an 'out' event
else if (events[i].second == 2)
{
// remove the 'y' coordinate from the active region
update(events[i].first.y, -1);
}
// if the current point is on a 'vertical' line
else
{
// find the range to be queried
int low = events[i++].first.y;
int up = events[i].first.y;
intersection_pts += query(up) - query(low);
}
}
return intersection_pts;
}
// returns (intersection_pts)C3
int findNumberOfTriangles()
{
int pts = findIntersectionPoints();
if ( pts >= 3 )
return ( pts * (pts - 1) * (pts - 2) ) / 6;
else
return 0;
}
// driver code
int main()
{
insertLine(point(2, 1), point(2, 9));
insertLine(point(1, 7), point(6, 7));
insertLine(point(5, 2), point(5, 8));
insertLine(point(3, 4), point(6, 4));
insertLine(point(4, 3), point(4, 5));
insertLine(point(7, 6), point(9, 6));
insertLine(point(8, 2), point(8, 5));
// sort the points based on x coordinate
// and event they are on
sort(events.begin(), events.end(), cmp);
cout << "Number of triangles are: " <<
findNumberOfTriangles() << "\n";
return 0;
}
输出：
Number of triangles are: 4
Time Complexity: O( n * log(n) + n * log(maximum_y) )