POJ 3304 Segments（判断直线与线段是否相交）

http://poj.org/problem?id=3304

等价于是否存在一条直线穿过所有线段

判断直线与线段是否相交：

首先用两点p1,p2确定了一条直线 在用p1,p2分别与计算线段两个端点计算叉乘即可 ，叉乘之积>0就说明线段两端点在直线的同侧，也就是直线不经过此线段

注意顺序不要搞反了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const double eps=1e-8;
const int maxn=100004;
int dcmp(double x) {
    if(fabs(x) < eps) return 0;
    return x < 0 ? -1 : 1;
}
#define Point Vector
struct Vector{
    double x,y;
    Vector(double x=0,double y=0):x(x),y(y){};
    Vector operator + (const Vector &A) const{
        return Vector(x+A.x,y+A.y);
    }
    Vector operator - (const Vector &A) const{
        return Vector(x-A.x,y-A.y);
    }
    Vector operator * (const double &A) const{
        return Vector(x*A,y*A);
    }
    Vector operator / (const double &A) const{
        return Vector(x/A,y/A);
    }
    bool operator == (const Vector &A) const{
        return dcmp(x-A.x)==0&&dcmp(y-A.y)==0;
    }
};
struct Line{
    Point p1,p2;
    Line(){};
    Line(Point p1,Point p2):p1(p1),p2(p2){}
}li[maxn];
bool operator < (Line l1,Line l2){
    if(l1.p1.x==l2.p1.x) return l1.p1.y<l2.p1.y;
    else return l1.p1.x<l2.p1.x;
}
int ans[maxn],num[maxn],n,m;
double U,L,R,D;
double PolarAngle(Vector A){
    return atan2(A.y,A.x);
}
Vector rotate(Vector &A,double a){
    return A=Vector(A.x*cos(a)-A.y*sin(a),A.x*sin(a)+A.y*cos(a));
}
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 dis(Point p1,Point p2){
    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
bool solve(Point p1,Point p2){
    if(dcmp(dis(p1,p2))==0) return false;
    for(int i=1;i<=n;i++){
        if(dcmp(Cross(li[i].p1-p1,li[i].p1-p2)*Cross(li[i].p2-p1,li[i].p2-p2))>0) return false;
    }
    return true;
}
int main(){
    int T;scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            Point p1,p2;
            scanf("%lf%lf%lf%lf",&p1.x,&p1.y,&p2.x,&p2.y);
            li[i]=Line(p1,p2);
        }
        if(n==1) {printf("Yes!\n");continue;}
        int flag=0;
        for(int i=1;i<=n;i++){
            if(solve(li[i].p1,li[i].p2)) {
                //cout<<i<<endl;
                printf("Yes!\n");flag=1;
                break;
            }
            for(int j=i+1;j<=n;j++){
                if(solve(li[i].p1,li[j].p1)) {
                    printf("Yes!\n");flag=1;
                    break;
                }
                if(solve(li[i].p1,li[j].p2)) {
                    printf("Yes!\n");flag=1;
                    break;
                }
                if(solve(li[i].p2,li[j].p1)) {
                    printf("Yes!\n");flag=1;
                    break;
                }
                if(solve(li[i].p2,li[j].p2)) {
                    printf("Yes!\n");flag=1;
                    break;
                }
            }
            if(flag) break;
        }
        if(!flag) printf("No!\n");
    }
}

原文地址：https://www.cnblogs.com/wifimonster/p/10323946.html