You can Solve a Geometry Problem too You can Solve a Geometry Problem too

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概述

You can Solve a Geometry Problem too

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 3 Accepted Submission(s) : 3

Problem Description

Many geometry（几何）problems were designed in the ACM/ICPC. And now, I also prepare a geometry problem for this final exam. According to the experience of many ACMers, geometry problems are always much trouble, but this problem is very easy, after all we are now attending an exam, not a contest :) Give you N (1<=N<=100) segments（线段）, please output the number of all intersections（交点）. You should count repeatedly if M (M>2) segments intersect at the same point. Note: You can assume that two segments would not intersect at more than one point.

Input

Input contains multiple test cases. Each test case contains a integer N (1=N<=100) in a line first, and then N lines follow. Each line describes one segment with four float values x1, y1, x2, y2 which are coordinates of the segment’s ending. A test case starting with 0 terminates the input and this test case is not to be processed.

Output

For each case, print the number of intersections, and one line one case.

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

Author

lcy

思路：排斥试验和跨立试验

代码如下：

#include <iostream>
#include <cstdio>
using namespace std;
typedef struct
{
double x , y;
}point;
typedef struct
{
point p1 , p2;
}line;
double xmulti(point p1 , line l1)
{
return (p1.x-l1.p1.x)*(l1.p2.y-l1.p1.y) - (l1.p2.x-l1.p1.x)*(p1.y-l1.p1.y);
}
bool inter(line l1 , line l2)
{
if(min(l1.p1.x , l1.p2.x) <= max(l2.p1.x , l2.p2.x)&&
min(l2.p1.x , l2.p2.x) <= max(l1.p1.x , l1.p2.x)&&
min(l1.p1.y , l1.p2.y) <= max(l2.p1.y , l2.p2.y)&&
min(l2.p1.y , l2.p2.y) <= max(l1.p1.y , l1.p2.y)&&
xmulti(l1.p1 , l2)*xmulti(l1.p2,l2) <= 0&&
xmulti(l2.p1 , l1)*xmulti(l2.p2 ,l1)<=0)
return true;
else
return false;
}
int main()
{
int n;
line l[105];
while(scanf("%d",&n),n)
{
for(int i = 0; i < n; i ++)
{
scanf("%lf%lf%lf%lf",&l[i].p1.x,&l[i].p1.y,&l[i].p2.x,&l[i].p2.y);
}
int c = 0;
for(int i = 0; i < n; i ++)
{
for(int j = i+1; j < n; j ++)
{
if(inter(l[i] , l[j]))
c ++;
}
}
printf("%dn",c);
}
return 0 ;
}