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[105];

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 0;
    for(int i=1;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 0;
        }
    }
    return 1;
}

bool ac(int n){
    Line tmp;
    int co=0;
    for(int i=1;i<=n;i++){
        for(int j=1;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=1;
            }
            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=1;
            }
            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=1;
            }
            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=1;
            }
        }

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

int main(){

    int T;
    cin>>T;
    while(T--){
        int n;
        cin>>n;
        Point s,e;
        for(int i=1;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!");
        }
    }

}