hdu_1086 You can Solve a Geometry Problem too(计算几何)
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http://acm.hdu.edu.cn/showproblem.php?pid=1086
分析:简单计算几何题,相交判断直接用模板即可。
思路:将第k条直线与前面k-1条直线进行相交判断,因为题目中不排除多条直线相交于同一个点的重复情况。
代码:
#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <queue>#include <map>#include <vector>#include <set>#include <string>#include <math.h>using namespace std;const double eps = 1e-8;const double PI = acos(-1.0); //比直接写3.1415926精确int sgn(double x) //三态函数,精确度提高{ if(fabs(x) < eps)return 0; else return x<0? -1:1;}struct Point{ double x,y;Point(){} Point(double _x,double _y) //带参构造函数 { x = _x;y = _y; } Point operator -(const Point &b)const //点相减 { return Point(x - b.x,y - b.y); } double operator ^(const Point &b)const //叉积(外积) { return x*b.y - y*b.x; } double operator *(const Point &b)const //点积 { return x*b.x + y*b.y; } void transXY(double B) //绕原点旋转角度B(弧度值),后x,y的变化 { double tx = x,ty = y; // x = tx*cos(B) - ty*sin(B); y = tx*sin(B) + ty*cos(B); }};struct Line{ Point s,e; Line(){} Line(Point _s,Point _e) { s = _s;e = _e; } //两直线相交求交点 //第一个值为0表示直线重合,为1表示平行,为0表示相交,为2是相交 //只有第一个值为2时,交点才有意义 pair<int,Point> operator &(const Line &b)const { Point res = s; if(sgn((s-e)^(b.s-b.e)) == 0) { if(sgn((s-b.e)^(b.s-b.e)) == 0) return make_pair(0,res);//重合 else return make_pair(1,res);//平行 } double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e)); res.x += (e.x-s.x)*t; res.y += (e.y-s.y)*t; return make_pair(2,res); }};Point ms,me;Line ml[120];//*两点间距离double dist(Point a,Point b){ return sqrt((a-b)*(a-b));}//*判断线段相交bool inter(Line l1,Line l2){ return max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) && max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) && max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) && max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) && sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e)) <= 0 && sgn((l1.s-l2.e)^(l2.s-l2.e))*sgn((l1.e-l2.e)^(l2.s-l2.e)) <= 0;}int main(){ int n; int ans; while(scanf("%d",&n),n){ ans=0; for(int i=0;i<n;i++){ scanf("%lf%lf%lf%lf",&ms.x,&ms.y,&me.x,&me.y); ml[i].s=ms; ml[i].e=me; } for(int i=1;i<n;i++) for(int j=0;j<i;j++){ if(inter(ml[i],ml[j])) ans++; } printf("%d\n",ans); } return 0;}
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