poj3304(叉乘)

有n條線段，問是否存在一條線段與所有給出得線段相交

存在這樣一條線段，那麼一定可以通過旋轉使得該線段與給出線段的某兩個端點相交，所以枚舉這樣的線段只需要枚舉任意兩個端點即可，判斷線段與線段是否相交，可以轉化成判斷線段1的兩個端點是否在線段2的兩側，然後用叉乘來判斷就行了，n注意==1時特判

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>

using namespace std;
const int mx = 105;
const double eps = 1e-8;
int n;

struct node {
	double x1, y1, x2, y2;
}segment[mx];

double dis(double x1, double y1, double x2, double y2) {
    return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));
}

double cross(double x1, double y1, double x2, double y2, double x, double y) {
    return (x1-x)*(y2-y) - (x2-x)*(y1-y);
}

int judge(double x1, double y1, double x2, double y2) {
    if (dis(x1,y1,x2,y2) <= eps) return 0;
    for (int i = 1; i <= n; i++) {
        if (cross(x1,y1,x2,y2,segment[i].x1,segment[i].y1) * cross(x1,y1,x2,y2,segment[i].x2,segment[i].y2) > eps)
            return 0;
    }
    return 1;
}

int main() {
	int T;
	scanf("%d",&T);

	while (T--) {
		scanf("%d",&n);
		for (int i = 1; i <= n; i++) {
			scanf("%lf%lf%lf%lf",&segment[i].x1,&segment[i].y1,&segment[i].x2,&segment[i].y2);
		}

		int flag = 0;
		for (int i = 1; i <= n; i++) {
			for (int j = i+1; j <= n; j++) {
				if (judge(segment[i].x1,segment[i].y1,segment[j].x1,segment[j].y1) ||
                    judge(segment[i].x1,segment[i].y1,segment[j].x2,segment[j].y2) ||
                    judge(segment[i].x2,segment[i].y2,segment[j].x1,segment[j].y1) ||
                    judge(segment[i].x2,segment[i].y2,segment[j].x2,segment[j].y2))
                    flag = 1;
			}
		}
		puts((flag||n==1)?"Yes!":"No!");
	}
	return 0;
}