题目链接：http://poj.org/PRoblem?id=1269 题意：给你两条线段的两个端点，让你判断他们的情况，总共有三种情况，共线（NONE），平行(LINE)，和相交(Point) 解析：共线和平行直接用斜率和叉乘来判断，剩下的就是求直线交点(推直线交点的时候心态别崩)
#include <cmath>#include <algorithm>#include <iostream>#include <cstdio>#include <vector>#include <cstring>#include <stack>using namespace std;const int maxn = 105+100;const double eps = 1e-8;struct point{ double x; double y; point() {} point(double _x,double _y) { x = _x; y = _y; }};struct line{ point a; point b;};double x_mul(point p0,point p1,point p2){ return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}bool judge1(line l1,line l2){ //判断是否平行 double t1 = (l2.a.x-l2.b.x)*(l1.a.y-l1.b.y); double t2 = (l2.a.y-l2.b.y)*(l1.a.x-l1.b.x); return t1==t2;}//是否相交//bool judge2(line l1,line l2)//{// bool flag1 = min(l1.a.x,l1.b.x)<=max(l2.a.x,l2.b.x)&&// min(l1.a.y,l1.b.y)<=max(l2.a.y,l2.b.y)&&// min(l2.a.x,l2.b.x)<=max(l1.a.x,l1.b.x)&&// min(l2.a.y,l2.b.y)<=max(l1.a.y,l1.b.y);// bool flag2 = (x_mul(l1.a,l2.a,l2.b)*x_mul(l1.b,l2.a,l2.b)<=0)&&// (x_mul(l2.a,l1.a,l1.b)*x_mul(l2.b,l1.a,l1.b)<=0);// return flag1&&flag2;//}point cross(line l1,line l2){ point res; double t1 = x_mul(l1.a,l1.b,l2.a); double t2 = x_mul(l1.b,l1.a,l2.b); res.x = (l2.b.x*t1+l2.a.x*t2)/(t1+t2); res.y = (l2.b.y*t1+l2.a.y*t2)/(t1+t2); return res;}int main(void){ int n; cin>>n; puts("INTERSECTING LINES OUTPUT"); while(n--) { line l1,l2; scanf("%lf %lf %lf %lf",&l1.a.x,&l1.a.y,&l1.b.x,&l1.b.y); scanf("%lf %lf %lf %lf",&l2.a.x,&l2.a.y,&l2.b.x,&l2.b.y); if(judge1(l1,l2)) { if(x_mul(l1.a,l2.a,l2.b)) puts("NONE"); else puts("LINE"); } else { point ans = cross(l1,l2); printf("POINT %.2f %.2f/n",ans.x,ans.y); } } puts("END OF OUTPUT"); return 0;}新闻热点
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