HDU 1086 You can Solve a Geometry Problem too

计算几何基础问题，具体可以参考算法导论p577，确定两个线段是否相交

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

struct point{
    double x,y;
};

struct segment{
    point p1,p2;
};

double direction(point p1, point p2, point p3){
    return (p2.x-p1.x)*(p3.y-p1.y)-(p2.y-p1.y)*(p3.x-p1.x);
}

bool on_segment(point p1, point p2, point p3){
    if( min(p1.x,p2.x) <= p3.x && p3.x <= max(p1.x,p2.x) && min(p1.y,p2.y) <= p3.y && p3.y <= max(p1.y,p2.y) )
        return true;
    else return false;
}

bool segment_intersect(point p1,point p2,point p3,point p4){
    double d1 = direction(p3,p4,p1);
    double d2 = direction(p3,p4,p2);
    double d3 = direction(p1,p2,p3);
    double d4 = direction(p1,p2,p4);
    if( d1*d2 <0 && d3*d4 < 0) return true;
    else if(d1 == 0 && on_segment(p3,p4,p1)) return true;
    else if(d2 == 0 && on_segment(p3,p4,p2)) return true;
    else if(d3 == 0 && on_segment(p1,p2,p3)) return true;
    else if(d4 == 0 && on_segment(p1,p2,p4)) return true;
    else return false;
}

int main(){
    int N;
    while(cin >>N && N){
        vector<segment> seg(N);
        for(int i = 0; i < N; i ++ )
            cin >> seg[i].p1.x>>seg[i].p1.y>>seg[i].p2.x>>seg[i].p2.y;
        int cnt  = 0;
        for(int i = 0 ; i < N; i ++ ){
            for(int j = i+1; j < N; j ++ ){
                if(segment_intersect(seg[i].p1,seg[i].p2,seg[j].p1,seg[j].p2)) cnt++;
            }
        }
        cout<< cnt<<endl;
    }
    return 0;
}