hdu 1086 You can Solve a Geometry Problem too

求出线段相交的对数

 

啦啦~~看算导的~~~我觉得计算几何蛮有意思的哇~~~不解释啦~~

 

#include <queue>#include <stack>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <iostream>#include <limits.h>#include <string.h>#include <algorithm>using namespace std;typedef struct Point{double x,y;}Point;typedef struct Line{Point a,b;}Line;double Direction(Point a, Point b, Point c ){return (c.x - a.x)*(b.y - a.y) - (b.x - a.x)*(c.y - a.y);}bool Onsegment(Point a, Point b, Point c){double maxx = max(a.x,b.x);double maxy = max(a.y,b.y);double minx = min(a.x,b.x);double miny = min(a.y,b.y);if( c.x >= minx && c.x <= maxx && c.y >= miny && c.y <= maxy )return true;return false;}bool segIntersect(Point p1,Point p2, Point p3, Point p4){double d1 = Direction(p3,p4,p1);double d2 = Direction(p3,p4,p2);double d3 = Direction(p1,p2,p3);double d4 = Direction(p1,p2,p4);if( d1 * d2 < 0 && d3 * d4 < 0 )return true;if( d1 == 0 && Onsegment(p3,p4,p1) )return true;if( d2 == 0 && Onsegment(p3,p4,p2) )return true;if( d3 == 0 && Onsegment(p1,p2,p3) )return true;if( d4 == 0 && Onsegment(p1,p2,p4) )return true;return false;}Line l[110];int main(){int n;int sum;while( ~scanf("%d",&n) && n ){sum = 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 i=0; i<n; i++)for(int k=i+1; k<n; k++)if( segIntersect(l[i].a,l[i].b,l[k].a,l[k].b) )sum++;printf("%d/n",sum);}return 0;}