51nod 判断线段是否相交 poj Segments直线与多条线段相交

给出两条线段的端点，判断是否相交

包括端点处的判断

若不包括端点处的，就在下面判断相交函数中去掉等号

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

const double eps=1e-8;
struct Point
{
double x,y;
Point(){}
Point(double _X, double _Y){
x = _X; y = _Y;
}
};

double Cross(Point p1,Point p2,Point p3){
return (p2.x-p1.x)*(p3.y-p1.y)-(p2.y-p1.y)*(p3.x-p1.x);
}
Point operator - (Point A,Point B){
return Point(A.x-B.x, A.y-B.y);
}
Point operator + (Point A, Point B){
return Point(A.x+B.x, A.y+B.y);
}
Point operator * (Point A, double p){
return Point(A.x*p, A.y*p);
}
bool operator == (Point A, Point B){
return (A.x-B.x) == 0 && (A.y-B.y) == 0;
}

bool SegmentProperIntersection(Point a,Point b,Point c,Point d)
{
return (max(a.x,b.x)>=min(c.x,d.x))&&
(max(c.x,d.x)>=min(a.x,b.x))&&
(max(a.y,b.y)>=min(c.y,d.y))&&
(max(c.y,d.y)>=min(a.y,b.y))&&
(Cross(a,c,b)*Cross(a,b,d)>=0)&&
(Cross(c,a,d)*Cross(c,d,b)>=0);
}

int main()
{
int T;
//freopen("in.txt","r",stdin);
scanf("%d",&T);
while(T--)
{
Point a,b,c,d;
scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);
scanf("%lf%lf%lf%lf",&c.x,&c.y,&d.x,&d.y);
int flag=SegmentProperIntersection(a,b,c,d);
puts(flag?"Yes":"No");
}
}

题意：

要你求是否存在一条直线，使得所有给出的线段投影在上面的影子至少有一个公共交点

题解：

由这一个公共点，我们可以推知，垂直这条线的直线，一定和所有线段相交

易知：极限情况下，这条直线一定过平面中某两个点

所以我们枚举两个点得到的直线是否和其他所有的线段都相交即可

#include<math.h>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define MAXN 110
const double eps=1e-8;
struct Point
{
double x,y;
Point(){}
Point(double _X,double _Y){
x = _X; y = _Y;
}
};

Point P[MAXN*2];

double Cross(Point p1,Point p2,Point p3){
return (p2.x-p1.x)*(p3.y-p1.y)-(p2.y-p1.y)*(p3.x-p1.x);
}
double Dis(Point p1,Point p2){
return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.x-p2.x)*(p1.x-p2.x));
}
Point operator - (Point A,Point B){
return Point(A.x-B.x, A.y-B.y);
}
Point operator + (Point A, Point B){
return Point(A.x+B.x, A.y+B.y);
}
Point operator * (Point A, double p){
return Point(A.x*p, A.y*p);
}
bool operator == (Point A, Point B){
return (A.x-B.x) == 0 && (A.y-B.y) == 0;
}

bool deal(int n)
{
int flag;
for(int i=1;i<=n*2;i++){
for(int j=i+1;j<=n*2;j++){
flag=1;
if(Dis(P[i],P[j])<eps)
continue;
for(int k=1;k<=n*2;k+=2){
if(Cross(P[i],P[j],P[k])*Cross(P[i],P[j],P[k+1])>0){
flag=0;
break;
}
}
if(flag)
return 1;
}
}
return 0;
}

int main()
{
int n,T;
//freopen("in.txt","r",stdin);
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=1;i<=n*2;i+=2)
scanf("%lf%lf%lf%lf",&P[i].x,&P[i].y,&P[i+1].x,&P[i+1].y);
if(deal(n))
puts("Yes!");
else
puts("No!");
}
return 0;
}