hdu 1086(判断线段相交)

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/*

*  题目要求：求一组线段的总的交点数

*  判断两线段相交：1快速排斥，2跨立实验  

*/

#include <cstdio>

#include <cstdlib>

#include <iostream>



using namespace std;



const int N = 100;



struct point {//点结构 

    double x;

    double y;

};



struct segment {//线段结构 

    point s;

    point e;

}seg[N];



double crossProd(point A, point B, point C) {

    return (B.x-A.x)*(C.y-A.y) - (B.y-A.y)*(C.x-A.x);

}



int segIntersect(segment seg1, segment seg2) {//判断线段是否相交 

    if (max(seg1.s.x, seg1.e.x) >= min(seg2.s.x, seg2.e.x) &&//快速排斥实验 

        max(seg2.s.x, seg2.e.x) >= min(seg1.s.x, seg1.e.x) &&

        max(seg1.s.y, seg1.e.y) >= min(seg2.s.y, seg2.e.y) &&

        max(seg2.s.y, seg2.e.y) >= min(seg1.s.y, seg1.e.y) &&

        crossProd(seg2.s, seg2.e, seg1.s) * crossProd(seg2.s, seg1.e, seg2.e) >= 0 &&//跨立实验 

        crossProd(seg1.s, seg1.e, seg2.s) * crossProd(seg1.s, seg2.e, seg1.e) >= 0)

        return 1;//相交 

    return 0;

}



int solve(int n) {

    int c = 0;

    for (int i=0; i<n; ++i) {

        for (int j=i+1; j<n; ++j) c += segIntersect(seg[i], seg[j]);

    }

    return c;

}



int main() {

    int n;

    while (scanf("%d", &n), n) {

        for (int i=0; i<n; ++i) scanf ("%lf%lf%lf%lf", &seg[i].s.x, &seg[i].s.y, &seg[i].e.x, &seg[i].e.y);

        int c = solve(n);

        printf ("%d\n", c);

    }

    return 0;

}