[Intersection of line segments] HDU 1086

Calculate the number of intersection times of a bunch of line segments, even if the intersection is overlapped.

For more information, see http://dev.gameres.com/program/abstract/geometry.htm?

 

Sample Input

2

0.00 0.00 1.00 1.00

0.00 1.00 1.00 0.00

3

0.00 0.00 1.00 1.00

0.00 1.00 1.00 0.000

0.00 0.00 1.00 0.00

0

 

Sample Output

1

3

 

 

C ++ code

# Include <iostream>

# Include <fstream>

# Include <algorithm>

# Include <string>

# Include <set>

// # Include <map>

# Include <queue>

# Include <utility>

# Include <iomanip>

# Include <stack>

# Include <list>

# Include <vector>

# Include <cstdio>

# Include <cstdlib>

# Include <cstring>

# Include <cmath>

# Include <ctime>

# Include <ctype. h>

Using namespace std;

# Define L _ int64

 

Struct point {// point structure

Double x, y;

Point (double a = 0, double B = 0) {x = a, y = B ;}

};

 

Struct line {// line Segment Structure

Point s, e;

Line (point a, point B) {s = a, e = B ;}

Line (){}

} L [2, 105];

 

Double multi (point a, point B, point c) // cross product to determine the point-to-line relationship

{

Double x1, y1, x2, y2;

X1 = B. x-a. x;

Y1 = B. y-a. y;

X2 = c. x-B. x;

Y2 = c. y-B. y;

Return x1 * y2-x2 * y1;

}

 

Bool intersect (line a, line B) // determines whether the two line segments are intersecting.

{

If (max (a. s. x, a. e. x)> = min (B. s. x, B. e. x) & // fast Rejection Test

Max (B. s. x, B. e. x)> = min (a. s. x, a. e. x )&&

Max (a. s. y, a. e. y)> = min (B. s. y, B. e. y )&&

Max (B. s. y, B. e. y)> = min (a. s. y, a. e. y )&&

Multi (a. s, B. s, B. e) * multi (a. e, B. s, B. e) <= 0 & // cross-site test

Multi (B. s, a. s, a. e) * multi (B. e, a. s, a. e) <= 0)

Return true;

Return false;

}

 

Int main ()

{

Int n, I, res, j;

While (scanf ("% d", & n), n)

{

For (I = 0; I <n; I ++)

Scanf ("% lf", & l [I]. s. x, & l [I]. s. y, & l [I]. e. x, & l [I]. e. y );

Res = 0;

For (I = 0; I <n; I ++)

For (j = I + 1; j <n; j ++)

If (intersect (l [I], l [j])

Res ++;

Printf ("% d \ n", res );

}

Return 0;

}