hdu 1086

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8586    Accepted Submission(s): 4177

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

20.00 0.00 1.00 1.000.00 1.00 1.00 0.0030.00 0.00 1.00 1.000.00 1.00 1.00 0.0000.00 0.00 1.00 0.000

 

Sample Output

13

求多条线段共有几个交点  不计算重复的

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<algorithm>#include<string>#include<map>#define ll __int64#define minn 1e-8using namespace std;struct Point{    double x,y;}P[10010];struct Line{    Point a,b;}L[110];///------------alg 2------------//叉积double mult(Point a, Point b, Point c){    return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);}//aa, bb为一条线段两端点 cc, dd为另一条线段的两端点 相交返回true, 不相交返回falsebool intersect(Point aa, Point bb, Point cc, Point dd){    if ( max(aa.x, bb.x)<min(cc.x, dd.x) )    {        return false;    }    if ( max(aa.y, bb.y)<min(cc.y, dd.y) )    {        return false;    }    if ( max(cc.x, dd.x)<min(aa.x, bb.x) )    {        return false;    }    if ( max(cc.y, dd.y)<min(aa.y, bb.y) )    {        return false;    }    if ( mult(cc, bb, aa)*mult(bb, dd, aa)<0 )    {        return false;    }    if ( mult(aa, dd, cc)*mult(dd, bb, cc)<0 )    {        return false;    }    return true;}///------------alg 2------------double cross(double x1, double y1, double x2, double y2){    return x1 * y2 - x2 * y1;}void intersec(Point a, Point b, Point c, Point d, double &x, double &y){    double u = cross(d.x - a.x, d.y - a.y, b.x - a.x, b.y - a.y);    double v = cross(c.x - a.x, c.y - a.y, b.x - a.x, b.y - a.y);    double w = u - v;    x = (u * 1.0 * c.x - v * 1.0 * d.x) / (w * 1.0);    y = (u * 1.0 * c.y - v * 1.0 * d.y) / (w * 1.0);}int main(){    int n;    while(~scanf("%d",&n))    {        if(n==0) break;        int len=0;        for(int i=0;i<n;i++)        {            scanf("%lf%lf%lf%lf",&L[i].a.x,&L[i].a.y,&L[i].b.x,&L[i].b.y);            for(int j=0;j<i;j++)            {                if(intersect(L[i].a,L[i].b,L[j].a,L[j].b))                {                    double st,ed;                    bool flag=true ;                    intersec(L[i].a,L[i].b,L[j].a,L[j].b,st,ed);                    for(int k=0;k<len;k++)                    {                        if(abs(P[k].x-st)<minn&&abs(P[k].y-ed)<minn)                        {                            flag=false ;                            break;                        }                    }                    if(flag)                        P[len].x=st,P[len].y=ed,len++;                }            }        }        printf("%d\n",len);    }}