poj 3304线段与直线相交

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

Segments

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9449		Accepted: 2902

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!

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

简短的题目描述，但题目确实挺难的

别人的题解很详细，我就不献丑了，下面题解的连接：http://hi.baidu.com/cloudayc/item/243f3c4c2a687aaadf2a9fb7

证明：坐标系内若干线段，是否存在一条直线与每条线段都有交点

转成叉积计算线段和直线是否相交,注意重点

转自discuss上的证明：from hanjialong

首先题中的要求等价于：存在一条直线l和所有的线段都相交。
证明：若存在一条直线l和所有线段相交，作一条直线m和l垂直，则m就是题中要求的直线，所有线段投影的一个公共点即为垂足。（l和每条线段的交点沿l投影到m上的垂足处）
反过来，若存在m，所有线段在m上的投影有公共点，则过这点垂直于m作直线l，l一定和所有线段相交。

然后证存在l和所有线段相交等价于存在l过某两条线段的各一个端点且和所有线段相交。
充分性显然。必要性：若有l和所有线段相交，则可保持l和所有线段相交，左右平移l到和某一线段交于端点停止（“移不动了”）。然后绕这个交点旋转。也是转到“转不动了”（和另一线段交于其一个端点）为止。这样就找到了一个新的l。

于是本题可归结为枚举两两线段的各一个端点，连一条直线，再判断剩下的线段是否都和这条直线有交点。

 #include <stdio.h>
 #include <string.h>
 #include <stdlib.h>
 #include <iostream>
 #include <algorithm>
 #include <math.h>

 using namespace std;
 #define eps 1e-8
 #define MAXX 1010

 typedef struct point
 {
     double x;
     double y;
 }point;

 typedef struct line
 {
     point st;
     point ed;
 }line;

 point p[MAXX];
 line li[MAXX];

 bool  dy(double x,double y){ return x>y+eps; }
 bool  xy(double x,double y){ return x<y-eps; }
 bool dyd(double x,double y){ return x>y-eps; }
 bool xyd(double x,double y){ return x<y+eps; }
 bool  dd(double x,double y){ return fabs(x-y)<eps; }

 double crossProduct(point a,point b,point c)
 {
     return (c.x-a.x)*(b.y-a.y)-(c.y-a.y)*(b.x-a.x);
 }

 double dist(point a,point b)
 {
     return sqrt((b.x-a.x)*(b.x-a.x)+(b.y-a.y)*(b.y-a.y));
 }

 bool onSegment(point a,point b,point c)
 {
     double maxx=max(a.x,b.x);
     double maxy=max(a.y,b.y);
     double minx=min(a.x,b.x);
     double miny=min(a.y,b.y);
     if(dd(crossProduct(a,b,c),0.0) && xyd(c.x,maxx) && dyd(c.x,minx) && xyd(c.y,maxy) && dyd(c.y,miny))
         return true;
     return false;
 }

 bool segIntersect(point p1,point p2,point p3,point p4)
 {
     double d1=crossProduct(p3,p4,p1);
     double d2=crossProduct(p3,p4,p2);
     double d3=crossProduct(p1,p2,p3);
     double d4=crossProduct(p1,p2,p4);
     /*if(xy(d1*d2,0.0) && xy(d3*d4,0.0))
         return true;
     if(dd(d1,0.0) && onSegment(p3,p4,p1))
         return true;
     if(dd(d2,0.0) && onSegment(p3,p4,p2))
         return true;
     if(dd(d3,0.0) && onSegment(p1,p2,p3))
         return true;
     if(dd(d4,0.0) && onSegment(p1,p2,p4))
         return true;
     return false;*/
     return xyd(d1 * d2,0.0);
 }

 int main()
 {
     int n,m,i,j,t;
     //freopen("in.txt","r",stdin);
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&n);
         int sub=,suc=;
         for(i=; i<n; i++)
         {
             scanf("%lf%lf%lf%lf",&p[sub].x,&p[sub].y,&p[sub+].x,&p[sub+].y);
             sub += ;
             li[suc].st.x=p[sub-].x;
             li[suc].st.y=p[sub-].y;
             li[suc].ed.x=p[sub-].x;
             li[suc++].ed.y=p[sub-].y;
         }
         if(n ==  || n == )
         {
             printf("Yes!\n");
             continue;
         }
         bool flag=false;
         for(i=; i<sub; i++)
         {
             for(j=i+; j<sub; j++)
             {
                 int sum=;
                 if(dist(p[i],p[j]) < eps)
                     continue;
                 for(int k=; k<suc; k++)
                 {
                     if(!segIntersect(li[k].st,li[k].ed,p[i],p[j]))
                     {
                          break;
                     }
                     else
                         sum++;
                 }//printf("%d***",sum);//test
                 if(sum >= n)
                 {
                     flag=true;
                     goto end;
                 }
                 //else sum=0;
             }
         }
     end:;
         if(flag)printf("Yes!\n");
         else printf("No!\n");
     }
     return ;
 }