Segments(叉积)

Segments

http://poj.org/problem?id=3304

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18066		Accepted: 5680

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!

枚举每两个线段的端点，判断这些端点形成的直线是否与线段都有交点
还有，题目说如果两个点的距离小于1e-8就看成一个点，所以要考虑重点

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<string>
 #include<algorithm>
 #include<queue>
 #include<vector>
 #include<map>
 using namespace std;

 struct Point{
     double x,y;
 };

 struct Line{
     Point s,e;
 }line[];

 double Cross(double x0,double y0,double x1,double y1,double x2,double y2){
     return (x1-x0)*(y2-y0)-(y1-y0)*(x2-x0);
 }

 double distance(double x1,double y1,double x2,double y2){
     return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
 }

 bool Check(double x1,double y1,double x2,double y2,int n){
     if(distance(x1,y1,x2,y2)<0.00000001) return ;
     for(int i=;i<=n;i++){
         if(Cross(x1,y1,x2,y2,line[i].s.x,line[i].s.y)*Cross(x1,y1,x2,y2,line[i].e.x,line[i].e.y)>0.00000001){
             return ;
         }
     }
     return ;
 }

 bool ac(int n){
     Line tmp;
     int co=;
     for(int i=;i<=n;i++){
         for(int j=;j<=n;j++){
             tmp.s=line[i].s;
             tmp.e=line[j].s;
             if(Check(tmp.s.x,tmp.s.y,tmp.e.x,tmp.e.y,n)){
                 co=;
             }
             tmp.s=line[i].s;
             tmp.e=line[j].e;
             if(Check(tmp.s.x,tmp.s.y,tmp.e.x,tmp.e.y,n)){
                 co=;
             }
             tmp.s=line[i].e;
             tmp.e=line[j].s;
             if(Check(tmp.s.x,tmp.s.y,tmp.e.x,tmp.e.y,n)){
                 co=;
             }
             tmp.s=line[i].e;
             tmp.e=line[j].e;
             if(Check(tmp.s.x,tmp.s.y,tmp.e.x,tmp.e.y,n)){
                 co=;
             }
         }

     }
     if(co)
         return true;
     return false;
 }

 int main(){

     int T;
     cin>>T;
     while(T--){
         int n;
         cin>>n;
         Point s,e;
         for(int i=;i<=n;i++){
             cin>>s.x>>s.y>>e.x>>e.y;
             line[i].s=s;
             line[i].e=e;
         }
         if(ac(n)){
             puts("Yes!");
         }
         else{
             puts("No!");
         }
     }

 }