POJ 1066 线段相交判断

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题目链接：http://poj.org/problem?id=1066

题解：线段相交判断

要通过的“门”最少，也就是该点与边框点的连线也其他线段的交点最少。

#include<cstdio>#include<cmath>#include<algorithm>#define N 35using namespace std;//定义点struct Point{double x, y;Point(double x = 0, double y = 0) : x(x), y(y) {}};typedef Point Vector; //Vector 为 Point的别名//点-点=向量Vector operator - (Point A, Point B) {return Vector(A.x-B.x, A.y-B.y);}struct segment{    Point a, b;}seg[N];const double eps = 1e-10;int dcmp(double x){if(fabs(x) < eps) return 0;else return x < 0 ? -1 : 1;}//叉积：两向量v和w的叉积等于v和w组成的三角形的有向面积的两倍 XaYb - XbYadouble Cross(Vector A, Vector B){return A.x*B.y - A.y*B.x;}//每条线段的两个端点都在另一条线段的两侧bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2)//a,b线段{double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1),   c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1);return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4) < 0;}int main (){    int n;    scanf("%d", &n);    for(int i = 1; i <= n; i++)    {        scanf("%lf %lf %lf %lf", &seg[i].a.x, &seg[i].a.y, &seg[i].b.x, &seg[i].b.y);    }    Point final;    scanf("%lf %lf", &final.x, &final.y);    int ans = 1000;    for(int i = 1; i <= n; i++)    {        int cnt = 0;        for(int j = 1; j <= n; j++)            if(SegmentProperIntersection(seg[j].a, seg[j].b, final, seg[i].a))                cnt++;        ans = min(cnt, ans);        cnt = 0;        for(int j = 1; j <= n; j++)            if(SegmentProperIntersection(seg[j].a, seg[j].b, final, seg[i].b))                cnt++;        ans = min(cnt, ans);    }    if(n == 0) printf("Number of doors = 1\n");     else    printf("Number of doors = %d\n", ans+1);    return 0;}

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