hdu 1086 You can Solve a Geometry Problem too 求n条直线交点的个数

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13549    Accepted Submission(s): 6645

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

 

题意：求n条直线交点的个数

 

#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
#include<math.h>
#include<string>
#include<string.h>
#include<vector>
#include<utility>
#include<map>
#include<queue>
#include<set>
#define mx 0x3f3f3f3f
#define ll long long
using namespace std;
const int N = 100010;
int flag;
double ans1,ans2,yy;
struct Point//定义点的结构体
{
    double x, y;
};
struct stline//定义边的结构体
{
    Point a, b;
} line[105];

bool cmp(Point a, Point b)
{
    return a.y < b.y;
}
int dblcmp(double a, double b)
{
    if (fabs(a - b) <= 1E-6) return 0;
    if (a > b) return 1;
    else return -1;
}
//***************点积判点是否在线段上***************
double dot(double x1, double y1, double x2, double y2) //点积
{
    return x1 * x2 + y1 * y2;
}

int point_on_line(Point a, Point b, Point c) //求a点是不是在线段bc上，>0不在，=0与端点重合，<0在。
{
    return dblcmp(dot(b.x - a.x, b.y - a.y, c.x - a.x, c.y - a.y), 0);
}
//**************************************************
double cross(double x1, double y1, double x2, double y2)
{
    return x1 * y2 - x2 * y1;
}
double ab_cross_ac(Point a, Point b, Point c) //ab与ac的叉积
{
    return cross(b.x - a.x, b.y - a.y, c.x - a.x, c.y - a.y);
}
int ab_cross_cd (Point a,Point b,Point c,Point d) //求ab是否与cd相交，交点为p。1规范相交，0交点是一线段的端点，-1不相交。
{
    double s1,s2,s3,s4;
    int d1,d2,d3,d4;
    Point p;
    d1=dblcmp(s1=ab_cross_ac(a,b,c),0);
    d2=dblcmp(s2=ab_cross_ac(a,b,d),0);
    d3=dblcmp(s3=ab_cross_ac(c,d,a),0);
    d4=dblcmp(s4=ab_cross_ac(c,d,b),0);

//如果规范相交则求交点
    if ((d1^d2)==-2 && (d3^d4)==-2)
    {
        p.x=(c.x*s2-d.x*s1)/(s2-s1);
        p.y=(c.y*s2-d.y*s1)/(s2-s1);
        return 1;
    }

//如果不规范相交
    if (d1==0 && point_on_line(c,a,b)<=0)
    {
        p=c;
        return 0;
    }
    if (d2==0 && point_on_line(d,a,b)<=0)
    {
        p=d;
        return 0;
    }
    if (d3==0 && point_on_line(a,c,d)<=0)
    {
        p=a;
        return 0;
    }
    if (d4==0 && point_on_line(b,c,d)<=0)
    {
        p=b;
        return 0;
    }
//如果不相交
    return -1;
}
int main()
{
    int t;
    while(scanf("%d", &t)&&t)
    {
        int cnt=0;
        for(int i=1;i<=t;i++)
            scanf("%lf%lf%lf%lf", &line[i].a, &line[i].a.y, &line[i].b.x, &line[i].b.y);
        for(int i=1;i<=t;i++)
        {
            for(int j=i+1;j<=t;j++)
            {
                if(ab_cross_cd(line[i].a, line[i].b, line[j].a, line[j].b)!=-1)
                    cnt++;
            }
        }
        printf("%d
",cnt);
    }
    return 0;
}