HDU 1086 You can Solve a Geometry Problem too
计算几何基础问题，具体可以参考算法导论p577，确定两个线段是否相交
1 #include <iostream> 2 #include <vector> 3 #include <algorithm> 4 5 using namespace std; 6 7 struct point{ 8 double x,y; 9 }; 10 11 struct segment{ 12 point p1,p2; 13 }; 14 15 double direction(point p1, point p2, point p3){ 16 return (p2.x-p1.x)*(p3.y-p1.y)-(p2.y-p1.y)*(p3.x-p1.x); 17 } 18 19 bool on_segment(point p1, point p2, point p3){ 20 if( min(p1.x,p2.x) <= p3.x && p3.x <= max(p1.x,p2.x) && min(p1.y,p2.y) <= p3.y && p3.y <= max(p1.y,p2.y) ) 21 return true; 22 else return false; 23 } 24 25 bool segment_intersect(point p1,point p2,point p3,point p4){ 26 double d1 = direction(p3,p4,p1); 27 double d2 = direction(p3,p4,p2); 28 double d3 = direction(p1,p2,p3); 29 double d4 = direction(p1,p2,p4); 30 if( d1*d2 <0 && d3*d4 < 0) return true; 31 else if(d1 == 0 && on_segment(p3,p4,p1)) return true; 32 else if(d2 == 0 && on_segment(p3,p4,p2)) return true; 33 else if(d3 == 0 && on_segment(p1,p2,p3)) return true; 34 else if(d4 == 0 && on_segment(p1,p2,p4)) return true; 35 else return false; 36 } 37 38 int main(){ 39 int N; 40 while(cin >>N && N){ 41 vector<segment> seg(N); 42 for(int i = 0; i < N; i ++ ) 43 cin >> seg[i].p1.x>>seg[i].p1.y>>seg[i].p2.x>>seg[i].p2.y; 44 int cnt = 0; 45 for(int i = 0 ; i < N; i ++ ){ 46 for(int j = i+1; j < N; j ++ ){ 47 if(segment_intersect(seg[i].p1,seg[i].p2,seg[j].p1,seg[j].p2)) cnt++; 48 } 49 } 50 cout<< cnt<<endl; 51 } 52 return 0; 53 }
