hdu/hdoj 1086 You can Solve a Geometry Problem too

計算兩條線段是否相交

推薦計算幾何文章http://dev.gameres.com/Program/Abstract/Geometry.htm#判斷兩線段是否相交

#include <iostream>  
using namespace std;  
 
struct line 
{  
    double x1,y1,x2,y2;  
}ln[105];  
 
int n;
 
int mul(double a, double b, double c, double d)  
{  
    return a * d - b * c;  
}
 
bool judge(int i, int j)
{
    double a = mul(ln[i].x1-ln[j].x1, ln[i].y1-ln[j].y1, ln[j].x2-ln[j].x1, ln[j].y2-ln[j].y1);
    double b = mul(ln[i].x2-ln[j].x1, ln[i].y2-ln[j].y1, ln[j].x2-ln[j].x1, ln[j].y2-ln[j].y1);
    double c = a * b;
 
    double d = mul(ln[j].x1-ln[i].x1, ln[j].y1-ln[i].y1, ln[i].x2-ln[i].x1, ln[i].y2-ln[i].y1);
    double e = mul(ln[j].x2-ln[i].x1, ln[j].y2-ln[i].y1, ln[i].x2-ln[i].x1, ln[i].y2-ln[i].y1);
    double f = d * e;
 
    if(c <= 0 && f <= 0)
        return 1;
    else
        return 0;
}
 
 
int main()  
{  
 
    while(scanf("%d", &n) && n)  
    {  
        for(int i=1; i<=n; ++i)  
            scanf("%lf %lf %lf %lf", &ln[i].x1, &ln[i].y1, &ln[i].x2, &ln[i].y2);  
        int count=0;  
        for(int i=1; i<n; ++i)  
        {  
            for(int j=i+1; j<=n; ++j)  
                if(judge(i,j)) 
                    ++count;  
        }  
        cout << count << endl;  
    }  
    return 0;  
}