hdu 3304(直线与线段相交)

Segments

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!可以理解成如果直线合法，然后旋转的话总是会和某两条直线的端点相交，枚举端点。找到一条合法的。这题用规范相交就过了。

///判断直线与线段相交
///做法：枚举每两个端点,要是存在一条直线经过这两个端点并且和所有线段相交就OK，但是不能为重合点.
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
const double eps = 1e-8;
const int N = 105;
struct Point
{
    double x,y;
};
struct Line{
    Point a,b;
}line[N];
int n;

double cross(Point a,Point b,Point c){
    return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);
}
double dis(Point a,Point b){
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
bool isCross(Point a,Point b,Point c,Point d){
    if(cross(a,b,c)*cross(a,b,d)>0) return true;
    return false;
}
bool check(Point a,Point b){
    if(sqrt(dis(a,b))<eps) return false;//这里记得讨论
    for(int i=0;i<n;i++){
        if(isCross(a,b,line[i].a,line[i].b)){
            return false;
        }
    }
    return true;
}
int main()
{
    int tcase;
    scanf("%d",&tcase);
    while(tcase--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%lf%lf%lf%lf",&line[i].a.x,&line[i].a.y,&line[i].b.x,&line[i].b.y);
        }
         if(n==1||n==2) {
            printf("Yes!\n");
            continue;
        }
        bool flag=false;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(check(line[i].a,line[j].a)||check(line[i].b,line[j].b)||check(line[i].a,line[j].b)||check(line[i].b,line[j].a)){
                    flag = true;
                    break;
                }
            }
            if(flag) break;
        }
        if(flag) printf("Yes!\n");
        else printf("No!\n");
    }
    return 0;
}