poj 3304 Segments（叉积+直线和线段相交判断）

Description

Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments on it, all projected segments have at least one point in common.

Input

Input begins with a number T showing the number of test cases and then, T test cases follow. Each test case begins with a line containing a positive integer n ≤ 100 showing the number of segments. After that, n lines containing four real numbers x1 y1 x2 y2 follow, in which (x1, y1) and (x2, y2) are the coordinates of the two endpoints for one of the segments.

Output

For each test case, your program must output “Yes!”, if a line with desired property exists and must output “No!” otherwise. You must assume that two floating point numbers a and b are equal if |a - b| < 10-8.

Sample Input

3
2
1.0 2.0 3.0 4.0
4.0 5.0 6.0 7.0
3
0.0 0.0 0.0 1.0
0.0 1.0 0.0 2.0
1.0 1.0 2.0 1.0
3
0.0 0.0 0.0 1.0
0.0 2.0 0.0 3.0
1.0 1.0 2.0 1.0
Sample Output

Yes!
Yes!
No!

大致题意：告诉你n个线段，问是否存在一条直线，使得这n条线段在该直线上的投影存在至少一个公共点？

思路：假设存在，那么从该公共点出发做一条与该直线垂直的直线，这条直线必定与这些所有线段相交。所以我们就可以将问题转化为求是否存在一条直线与这些所有线段相交。n<=100，所以我们可以去枚举所有线段的两个端点假设为所求直线，然后用叉积去判断该直线是否与所有线段相交。
注意：可能会枚举到重合的点，特殊考虑下。

代码如下

#include<iostream>#include<set>#include<vector>#include<algorithm>#include<cstring>#include<cstdio>#include<cmath>using namespace std;typedef long long ll;const int N=105;const double eps=1e-8;int n;int dcmp(double x) {    if(fabs(x)<eps) return 0;    return x<0?-1:1;}struct Point {    double x,y;    Point() {}    Point(double  _x,double _y) {        x=_x;        y=_y;    }    Point operator-(const Point &b) const {        return Point(x-b.x,y-b.y);    }    double operator *(const Point &b)const {        return x*b.x + y*b.y;    }    double operator ^(const Point &b)const {        return x*b.y - y*b.x;    }};struct Line {    Point a,b;    Line() {}    Line(Point _a,Point _b) {        a=_a;        b=_b;    }};Line line[N];double dist(Point a,Point b){    return sqrt((b-a)*(b-a));//返回两个点之间的距离   } double xmult(Point p0,Point p1,Point p2) {    return (p1-p0)^(p2-p0);}bool check(Line L1,Line L2)//判断直线L1和线段L2是否相交 {    return dcmp(xmult(L2.a,L1.a,L1.b))*dcmp(xmult(L2.b,L1.a,L1.b))<=0;}bool solve(Line L1){    if(dcmp(dist(L1.a,L1.b))==0) return false;//如果所选的两个点重合,即不能构成一条直线     for(int i=0;i<n;i++)        if(check(L1,line[i])==false)        return false;    return true; }int main() {    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        double x1,y1,x2,y2;        for(int i=0;i<n;i++)        {            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);            line[i]=Line(Point(x1,y1),Point(x2,y2));        }        int f=0;        for(int i=0;i<n;i++)        if(!f)        for(int j=0;j<n;j++)        {            if(solve(Line(line[i].a,line[j].a))||solve(Line(line[i].a,line[j].b))||            solve(Line(line[i].b,line[j].a))||solve(Line(line[i].b,line[j].b)))            {                f=1;                break;            }        }        if(f)            printf("Yes!\n");        else            printf("No!\n");    }     return 0;}