poj 3304 直线与线段相交

Segments

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12161		Accepted: 3847

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!

/*
poj 3304 直线与线段相交

给你n条线段,确定是否存在一条直线，它们的投影到上面时有公共点

如果存在一个这样的直线,那么说明所有的线段能和一条直线相交
对这条直线进行一定的旋转,必定与所有直线的端点中的至少两个相交
所以枚举所有的端点进行判断即可
//注意判断相同点

hhh-2016-05-04 20:48:26
*/
#include <iostream>
#include <vector>
#include <cstring>
#include <string>
#include <cstdio>
#include <queue>
#include <cmath>
#include <algorithm>
#include <functional>
#include <map>
using namespace std;
#define lson  (i<<1)
#define rson  ((i<<1)|1)
typedef long long ll;
const int  maxn = 40010;
double eps = 1e-8;
int tot;
int n,m;
double x1,x2,y1,y2;

int sgn(double x)
{
    if(fabs(x) < eps) return 0;
    if(x < 0)
        return -1;
    else
        return 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.y-y*b.x;
    }
};

struct Line
{
    Point s,t;
    Line() {}
    Line(Point _s,Point _t)
    {
        s = _s;
        t = _t;
    }
};
int tans[maxn];
Line line[maxn];
Point po[maxn];
Point p;

bool seg_inter_line(Line l1,Line l2)
{
    return sgn((l2.s-l1.t) ^ (l1.s-l1.t))*sgn((l2.t-l1.t)^(l1.s-l1.t)) <= 0;
}

bool cal(Line tl)
{
    for(int i = 0;i < n;i++)
    {
        if(!seg_inter_line(tl,line[i]))
        {
            return false;
        }
    }
    return true;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        int flag = 0;
        tot = 0;
        if(n < 3)
            flag = 1;
        for(int i = 0; i < n; i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            po[tot++] = Point(x1,y1);
            po[tot++] = Point(x2,y2);
            line[i] = Line(po[tot-1],po[tot-2]);
        }
        for(int i = 0; i < tot && !flag; i++)
        {
            for(int j = i+1; !flag && j < tot; j++)
            {
                if(po[i].x == po[j].x && po[i].y == po[j].y)
                    continue;
                if(cal(Line(po[i],po[j])))
                {
                    flag = 1;
                    break;
                }
            }
        }
        if(flag)
            printf("Yes!
");
        else
            printf("No!
");
    }
    return 0;
}