POJ 3304 Segments（判断直线和线段相交）

Segments

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12150 Accepted: 3845

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

321.0 2.0 3.0 4.04.0 5.0 6.0 7.030.0 0.0 0.0 1.00.0 1.0 0.0 2.01.0 1.0 2.0 1.030.0 0.0 0.0 1.00.0 2.0 0.0 3.01.0 1.0 2.0 1.0

Sample Output

Yes!Yes!No!

Source

Amirkabir University of Technology Local Contest 2006

原博：http://blog.csdn.net/wangjian8006

题目大意：给出n条线段两个端点的坐标，问所有线段投影到一条直线上，如果这些所有投影至少相交于一点就输出Yes！，否则输出No！。
解题思路：如果有存在这样的直线，过投影相交区域作直线的垂线，该垂线必定与每条线段相交，问题转化为问是否存在一条线和所有线段相交
若存在一条直线与所有线段相机相交，此时该直线必定经过这些线段的某两个端点，所以枚举任意两个端点即可。
这里要主要的地方就是，题目说如果两个点的距离小于1e-8就等价于一点，所以要考虑重点

/*Memory 200KTime   32MS */#include <iostream>#include <math.h>using namespace std;#define MAXM 110#define EPS 1e-8typedef struct{double x1,y1,x2,y2;}Segment;Segment segment[MAXM];int n;double distance(double x1,double y1,double x2,double y2){return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}double corss(double x1,double y1,double x2,double y2,double x,double y){return (x2-x1)*(y-y1)-(x-x1)*(y2-y1);}bool judge(double x1,double y1,double x2,double y2){int i;if(distance(x1,y1,x2,y2)<EPS) return 0;for(i=0;i<n;i++){if(corss(x1,y1,x2,y2,segment[i].x1,segment[i].y1)*corss(x1,y1,x2,y2,segment[i].x2,segment[i].y2)>EPS) return 0;}return 1;}int main(){int t,i,j,ans;scanf("%d",&t);while(t--){scanf("%d",&n);for(i=0;i<n;i++)scanf("%lf%lf%lf%lf",&segment[i].x1,&segment[i].y1,&segment[i].x2,&segment[i].y2);if(n==1) {printf("Yes!\n");continue;}ans=0;for(i=0;i<n && !ans;i++)for(j=i+1;j<n && !ans;j++){if(judge(segment[i].x1,segment[i].y1,segment[j].x1,segment[j].y1) ||judge(segment[i].x1,segment[i].y1,segment[j].x2,segment[j].y2) ||judge(segment[i].x2,segment[i].y2,segment[j].x1,segment[j].y1) ||judge(segment[i].x2,segment[i].y2,segment[j].x2,segment[j].y2))ans=1;}if(ans) printf("Yes!\n");else printf("No!\n");}return 0;