You can Solve a Geometry Problem too（线段求交）

http://acm.hdu.edu.cn/showproblem.php?pid=1086

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8861    Accepted Submission(s): 4317

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

 

题解：本题题干已经排除了两线重合的多边交于一点的情况，故直接枚举所有的边是否相交即可

#include<cstdio>
#include<cmath>
using namespace std;
#define eps 1e-6
#define N 105
struct point{
    double x , y ;
    point(double x_, double y_){
         x = x_;
         y = y_;
    }
    point(){}
    point operator - (const point a) const
    {
        return point(x-a.x,y-a.y);
    }
    double operator * (const point a) const
    {
        return x*a.y - a.x*y;
    }
};

struct line{
    point s , t;
}L[N];

int main()
{
    int T;
    while(~scanf("%d",&T),T)
    {
        for(int i = 0 ;i < T ; i++)
        {
            scanf("%lf%lf%lf%lf",&L[i].s.x,&L[i].s.y,&L[i].t.x,&L[i].t.y);
        }
        int ans = 0;
        for(int i = 0 ; i < T ; i++)
        {
            for(int j = i+1 ; j < T ; j++)//j从i开始保证不会重复判断
            {
               // if(i==j) continue;
                point A = L[i].s;
                point B = L[i].t;
                point C = L[j].s;
                point D = L[j].t;
                if((((D-C)*(A-C))*((D-C)*(B-C)))>eps) {continue;}
                if((((D-A)*(B-A))*((C-A)*(B-A)))>eps) {continue;}
                    ans++;
            }
        }
        printf("%d
",ans);
    }
    return 0;
}

也可以把他们写成函数在外面

#include <cstdio>
#include <cmath>
using namespace std;
#define eps 1e-8
#define N 105
struct point{
    double x, y;
    point(){}
    point(double _x, double _y) {
        x = _x, y = _y;
    }
    
    point operator - (point a){
        return point(x-a.x, y-a.y);
    }
    
    double operator * (point a){
        return x*a.y - y*a.x;
    }
};

struct line{
    point s, t;
}L[N];

bool ck(line a, line b)
{
    point A = a.s, B = a.t, C = b.s, D = b.t;
    if(((C-A)*(B-A)) *((D-A)*(B-A)) > eps) return false;
    if(((A-C)*(D-C)) *((B-C)*(D-C)) > eps) return false;
    return true;
}

int main()
{
    int n;
    while(~scanf("%d", &n), n)
    {
        for(int i = 0; i < n; i++)
            scanf("%lf %lf %lf %lf", &L[i].s.x, &L[i].s.y, &L[i].t.x, &L[i].t.y);
        int cnt = 0;
        for(int i = 0; i < n; i++)
            for(int j = i+1; j < n; j++)
                cnt += ck(L[i], L[j]);
        printf("%d
", cnt);
    }
}