题目:意思就是判断给定的几条线段是否有相交的。
方法:模版吧,有空在来细细学习。
代码:
#include <iostream> #include <cstdio> using namespace std; struct Point { double x,y; }; struct LineSeg { Point a,b; }; double Cross(Point a, Point b, Point c ) { return (c.x - a.x)*(b.y - a.y) - (b.x - a.x)*(c.y - a.y); } int Yu(LineSeg u,LineSeg v) { return( (max(u.a.x,u.b.x)>=min(v.a.x,v.b.x))&& (max(v.a.x,v.b.x)>=min(u.a.x,u.b.x))&& (max(u.a.y,u.b.y)>=min(v.a.y,v.b.y))&& (max(v.a.y,v.b.y)>=min(u.a.y,u.b.y))&& (Cross(v.a,u.b,u.a)*Cross(u.b,v.b,u.a)>=0)&& (Cross(u.a,v.b,v.a)*Cross(v.b,u.b,v.a)>=0)); } int main() { int n; int flag=0; LineSeg l[2002]; while(cin>>n) { flag=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); if(!flag) for(int j=0;j<i;j++) if(Yu(l[j],l[i])) { flag=1; break; } } if(!flag) cout<<"ok!"<<endl; else cout<<"burned!"<<endl; } return 0; }
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原文地址:http://blog.csdn.net/knight_kaka/article/details/25426933