POJ 3304 Segments

Segments

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6615		Accepted: 1981

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

 

计算几何，直线与线段的相交，

题意：要计算是否存在一条直线，所有给出的线段投影到这条直线上的时候，有一个公共的交点。

我们可以这样想，如果存在一条直线满足与所有的直线都有交点的话，那么这条直线的垂线就是符合题意的一条直线，因为在这条直线的每个交点上做一个垂线的话，这个方向就会满足题目的要求，

所以我们可以转为寻找是否存在一条与所有线段都相交的直线，

 

如果存在一条直线与所有的线段相交的话，那么，我们可以通过平移，让这条直线与其中的一条线段的端点重合，再进一步，我们也可以通过平移或转动，让这条直线与另外的一条线段的端点相交。这样，每一条符合题意的直线，必然与线段的两个端点重合

 

#include<stdio.h>
#include<cmath>
#include<iostream>
using namespace std;
#define eps 1e-8 
struct Point {
    double x,y ;
}p[410] ;
double multi(Point p0,Point p1,Point p2){
    return (p1.x-p0.x)*(p2.y-p0.y) - (p2.x-p0.x)*(p1.y-p0.y) ;
}
bool work(int n){
    int flag ;
    for(int i=1;i<=n;i++){
        for(int j=1+i;j<=n;j++){
            if(abs(p[j].x-p[i].x)>eps || abs(p[j].y-p[i].y)>eps){
                flag = 1;
                for(int k=1;k<=n;k+=2){
                    if(multi(p[i],p[j],p[k])*multi(p[i],p[j],p[k+1])> 0){
                        flag = 0;
                        break;
                    }
                }
                if(flag)     return 1;
            }
        }
    }
    return 0 ;
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int n;
        scanf("%d",&n) ;
        int cnt = 1;
        for(int i=1;i<=n;i++,cnt+=2)
            scanf("%lf%lf%lf%lf",&p[cnt].x,&p[cnt].y,&p[cnt+1].x,&p[cnt+1].y) ;    
        if(work(cnt-1))
            puts("Yes!") ;
        else
            puts("No!") ;
    }
    return 0;
}