POJ 3304：Segments 计算几何 是否有直线与所有线段相交

Segments

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15648		Accepted: 4971

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!

题意就是问 是否有一条直线

使得所有线段在该线段上的投影有公共点

之后把问题转化为是不是存在直线使得它与所有的线段都有交点（在该共同区域做垂线，一定可以交）

所以枚举每两个个线段的端点就好了

#include<cmath>
#include<ctime>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<string>
#include<bitset>
#include<queue>
#include<map>
#include<set>
using namespace std;

typedef double db;

const int N=1010;
const db eps=1e-8;

int n;
db a[N],b[N],c[N],d[N];

inline db xmul(db x,db y,db x2,db y2,db basx,db basy)
{return (x-basx)*(y2-basy)-(x2-basx)*(y-basy);}

inline bool check(db x,db y,db x2,db y2)
{
	if(abs(x-x2)<eps&&abs(y-y2)<eps)return 0;
	for(int i=1;i<=n;++i)
	if(xmul(a[i],b[i],x,y,x2,y2)*xmul(c[i],d[i],x,y,x2,y2)>eps)return 0;
	return 1;
}

int main()
{
	int T;
	register int i,j;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		for(i=1;i<=n;++i)scanf("%lf%lf%lf%lf",&a[i],&b[i],&c[i],&d[i]);
		bool flag=0;
		for(i=1;i<=n;++i)
		{
			for(j=1;j<=n;++j)
			{
				if(check(a[i],b[i],a[j],b[j])){flag=1;break;}
				if(check(a[i],b[i],c[j],d[j])){flag=1;break;}
				if(check(c[i],d[i],a[j],b[j])){flag=1;break;}
				if(check(c[i],d[i],c[j],d[j])){flag=1;break;}
			}
			if(flag)break;
		}
		flag?puts("Yes!"):puts("No!");
	}
	return 0;
}
/*
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

Yes!
Yes!
No!
*/