poj 1269 水题
题目链接:http://poj.org/problem?id=1269
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cstdio> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> #include<queue> using namespace std; const int maxn = 105; const int maxe = 20000; const int INF = 0x3f3f3f; const double eps = 1e-8; const double PI = acos(-1.0); struct Point{ double x,y; Point(double x=0, double y=0) : x(x),y(y){ } //构造函数 }; typedef Point Vector; Vector operator + (Vector A , Vector B){return Vector(A.x+B.x,A.y+B.y);} Vector operator - (Vector A , Vector B){return Vector(A.x-B.x,A.y-B.y);} Vector operator * (Vector A , double p){return Vector(A.x*p,A.y*p);} Vector operator / (Vector A , double p){return Vector(A.x/p,A.y/p);} bool operator < (const Point& a,const Point& b){ return a.x < b.x ||( a.x == b.x && a.y < b.y); } int dcmp(double x){ if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; } bool operator == (const Point& a, const Point& b){ return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0; } double Dot(Vector A, Vector B){ return A.x*B.x + A.y*B.y; } double Cross(Vector A, Vector B) { return A.x*B.y - A.y * B.x; } double Length(Vector A) { return sqrt(Dot(A,A)); } Point GetLineIntersecion(Point P, Vector v,Point Q,Vector w){ Vector u = P - Q; double t = Cross(w,u)/Cross(v,w); return P + v*t; } Point read_point(){ Point A; scanf("%lf %lf",&A.x,&A.y); return A; } /******************************分割线*******************************/ Point P1,P2,Q1,Q2; int N; int main() { // freopen("E:\\acm\\input.txt","r",stdin); cin>>N; printf("INTERSECTING LINES OUTPUT\n"); while(N--){ P1 = read_point(); P2 = read_point(); Q1 = read_point(); Q2 = read_point(); if(dcmp(Cross(P2-P1,Q2-Q1))==0){ //平行或重合; if(dcmp(Cross(P1-Q1,P1-Q2)) == 0) printf("LINE\n"); else printf("NONE\n"); } else{ //相交; Point A = GetLineIntersecion(P1,P2-P1,Q1,Q2-Q1); printf("POINT %.2f %.2f\n",A.x,A.y); } } printf("END OF OUTPUT\n"); }