You can Solve a Geometry Problem too
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12425 Accepted Submission(s): 6103
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
//模板
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const double eps = 1e-10;
struct point
{
double x, y;
};
struct Line{
point l;
point r;
};
double min(double a, double b)
{
return a < b ? a : b;
}
double max(double a, double b)
{
return a > b ? a : b;
}
bool inter(point a, point b, point c, point d)
{//a为一条直线的起点b为它的终点,c为另一条直线的起点d为这条直线的终点
if (min(a.x, b.x) > max(c.x, d.x) || min(a.y, b.y) > max(c.y, d.y) || min(c.x, d.x) > max(a.x, b.x) || min(c.y, d.y) > max(a.y, b.y))
{
return 0;
}
double h, i, j, k;
h = (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
i = (b.x - a.x) * (d.y - a.y) - (b.y - a.y) * (d.x - a.x);
j = (d.x - c.x) * (a.y - c.y) - (d.y - c.y) * (a.x - c.x);
k = (d.x - c.x) * (b.y - c.y) - (d.y - c.y) * (b.x - c.x);
return h * i <= eps && j * k <= eps;
}
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const double eps = 1e-10;
struct point
{
double x, y;
};
struct Line{
point l;
point r;
};
double min(double a, double b)
{
return a < b ? a : b;
}
double max(double a, double b)
{
return a > b ? a : b;
}
bool inter(point a, point b, point c, point d)
{//a为一条直线的起点b为它的终点,c为另一条直线的起点d为这条直线的终点
if (min(a.x, b.x) > max(c.x, d.x) || min(a.y, b.y) > max(c.y, d.y) || min(c.x, d.x) > max(a.x, b.x) || min(c.y, d.y) > max(a.y, b.y))
{
return 0;
}
double h, i, j, k;
h = (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
i = (b.x - a.x) * (d.y - a.y) - (b.y - a.y) * (d.x - a.x);
j = (d.x - c.x) * (a.y - c.y) - (d.y - c.y) * (a.x - c.x);
k = (d.x - c.x) * (b.y - c.y) - (d.y - c.y) * (b.x - c.x);
return h * i <= eps && j * k <= eps;
}
int main()
{
int n;
Line line[110];
while(cin>>n&&n)
{
for(int i=1;i<=n;i++)
{
cin>>line[i].l.x>>line[i].l.y>>line[i].r.x>>line[i].r.y;
}
int cnt=0;
for(int i=1;i<=n;i++)
{
for(int j=i+1;j<=n;j++)
{
if(inter(line[i].l,line[i].r,line[j].l,line[j].r))
cnt++;
}
}
cout<<cnt<<endl;
}
}