You can Solve a Geometry Problem too（判断两线段是否相交）

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 199    Accepted Submission(s): 132

Problem Description

Many geometry（几何）problems were designed in the ACM/ICPC. And now, I also prepare a geometry problem for this final exam. According to the experience of many ACMers, geometry problems are always much trouble, but this problem is very easy, after all we are now attending an exam, not a contest :) Give you N (1<=N<=100) segments（线段）, please output the number of all intersections（交点）. You should count repeatedly if M (M>2) segments intersect at the same point.
Note: You can assume that two segments would not intersect at more than one point.

 

Input

Input contains multiple test cases. Each test case contains a integer N (1=N<=100) in a line first, and then N lines follow. Each line describes one segment with four float values x1, y1, x2, y2 which are coordinates of the segment’s ending. A test case starting with 0 terminates the input and this test case is not to be processed.

 

Output

            For each case, print the number of intersections, and one line one case.

 

Sample Input

2
0.00 0.00 1.00 1.00
0.00 1.00 1.00 0.00
3
0.00 0.00 1.00 1.00
0.00 1.00 1.00 0.000
0.00 0.00 1.00 0.00
0

 

Sample Output

1
3

 

Author

lcy

 

/*
计算几何求所给线段的交点数量

*/
#include<bits/stdc++.h>
using namespace std;
struct Point{//点
    double x,y;
    Point(){}
    Point(int a,int b){
        x=a;
        y=b;
    }
    void input(){
        scanf("%lf%lf",&x,&y);
    }
};
struct Line{//线段
    Point a,b;
    Line(){}
    Line(Point x,Point y){
        a=x;
        b=y;
    }
    void input(){
        a.input();
        b.input();
    }
};
bool judge(Point &a,Point &b,Point &c,Point &d)
{
    if(!(min(a.x,b.x)<=max(c.x,d.x) && min(c.y,d.y)<=max(a.y,b.y)&&
        min(c.x,d.x)<=max(a.x,b.x) && min(a.y,b.y)<=max(c.y,d.y)))//这里的确如此，这一步是判定两矩形是否相交
        //特别要注意一个矩形含于另一个矩形之内的情况
    return false;
    double u,v,w,z;//分别记录两个向量
    u=(c.x-a.x)*(b.y-a.y)-(b.x-a.x)*(c.y-a.y);
    v=(d.x-a.x)*(b.y-a.y)-(b.x-a.x)*(d.y-a.y);
    w=(a.x-c.x)*(d.y-c.y)-(d.x-c.x)*(a.y-c.y);
    z=(b.x-c.x)*(d.y-c.y)-(d.x-c.x)*(b.y-c.y);
    return (u*v<=0.00000001 && w*z<=0.00000001);
}
vector<Line>v;//用来存放线段
int n;
Line a;
void init(){
    v.clear();
}
int main(){
    //freopen("in.txt","r",stdin);
    while(scanf("%d",&n)!=EOF&&n){
        init();
        for(int i=0;i<n;i++){
            a.input();
            v.push_back(a);
        }//将线段存入
        int cur=0;
        for(int i=0;i<v.size();i++){
            for(int j=i+1;j<v.size();j++)
                if(judge(v[i].a,v[i].b,v[j].a,v[j].b))
                    cur++;
        }
        printf("%d
",cur);
    }
    return 0;
}