Segments
Time Limit: 1000MS Memory Limit: 65536K
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!
Source
Amirkabir University of Technology Local Contest 2006
又是一道基础的计算几何题，就是询问是否存在一条直线穿过给定的所有线段，由于n" role="presentation" style="position: relative;">nn很小，我们直接暴力枚举两个端点表示直线然后再O（n）" role="presentation" style="position: relative;">O（n）O（n）判断就行了（本蒟蒻因为有个return" role="presentation" style="position: relative;">returnreturn没有写调了40min" role="presentation" style="position: relative;">40min40min）。
代码如下：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define eps 1e-8
#define N 105
using namespace std;
struct pot{double x,y;}p[N<<1];
int n,t;
inline int sign(double x){return (x>eps)-(x<-eps);}
inline pot operator-(pot a,pot b){return pot{a.x-b.x,a.y-b.y};}
inline double cross(pot a,pot b){return a.x*b.y-a.y*b.x;}
inline bool ok(pot a,pot b,pot c,pot d){
if((cross(a-c,b-c)*cross(a-d,b-d))<=0.0000)return true;
return false;
}
inline bool pd(pot a,pot b){
for(int i=1;i<n;i+=2)if(ok(a,b,p[i],p[i+1])==0)return false;
return true;
}
inline bool check(){
for(int i=1;i<n;++i)
for(int j=i+1;j<=n;++j){
if(sign(p[i].x-p[j].x)==0&&sign(p[i].y-p[j].y)==0)continue;
if(pd(p[i],p[j]))return true;
}
return false;
}
int main(){
scanf("%d",&t);
while(t--){
scanf("%d",&n);
n<<=1;
for(int i=1;i<=n;++i)scanf("%lf%lf",&p[i].x,&p[i].y);
if(check())puts("Yes!");
else puts("No!");
}
return 0;
}