poj 2826 计算几何

（代码后面给出了一些样例，都过了的话应该是没有）

乍一看就是个水题，实际上奥妙重重！！！

主要是由一种上面线段覆盖下面线段的情况，开始没注意到。

 这种情况显然输出0。

#include<stdio.h>#include<string.h>#include<ctype.h>#include<math.h>#include<string>#include<set>#include<map>#include<vector>#include<queue>#include<algorithm>#include<ctime>using namespace std;void fre(){freopen("t.txt","r",stdin);}template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }#define ls o<<1#define rs o<<1|1#define MS(x,y) memset(x,y,sizeof(x))#define debug(x) printf("%d",x);typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;const int INF = 1<<30;const int MAXN = 105;const double eps = 1e-8;const double pi = acos(-1.0);int sgn(double x){    if(fabs(x) < eps) return 0;    if(x < 0) return -1;    else return 1;}struct Point{    double x,y;    Point(){}    Point(double _x,double _y)    {        x = _x; y = _y;    }    void input()    {        scanf("%lf%lf",&x,&y);    }    Point operator -(const Point &b) const    {        return Point(x-b.x,y-b.y);    }    double operator ^(const Point &b) const    {        return x*b.y - y*b.x;    }    double operator *(const Point &b) const    {        return x*b.x + y*b.y;    }};struct Line{    Point s,e;    Line(){}    Line(Point _s,Point _e)    {        s = _s; e = _e;    }    double getk()    {        return (s.y-e.y)/(s.x-e.x);    }    bool parallel(Line v)    {        return sgn((e-s)^(v.e-v.s)) == 0;    }    void input()    {        e.input(); s.input();        if(s.y < e.y) swap(s,e);    }    int segcrossseg(Line v)    {        int d1 = sgn((e-s)^(v.s-s));        int d2 = sgn((e-s)^(v.e-s));        int d3 = sgn((v.e-v.s)^(s-v.s));        int d4 = sgn((v.e-v.s)^(e-v.s));        if( (d1^d2)==-2 && (d3^d4)==-2 ) return 2;        return (d1 == 0 && sgn((v.s-s)*(v.s-e)) <= 0) ||            (d2 == 0 && sgn((v.e-s)*(v.e-e)) <= 0) ||            (d3 == 0 && sgn((s-v.s)*(s-v.e)) <= 0) ||            (d4 == 0 && sgn((e-v.s)*(e-v.e)) <= 0);    }    Point crosspoint(Line v)    {        double a1 = (v.e-v.s)^(s-v.s);        double a2 = (v.e-v.s)^(e-v.s);        return Point( (s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));    }}l1,l2;int main(){    //fre();    int T;    double ans;    scanf("%d",&T);    while(T--)    {        l1.input(); l2.input();        if(l1.s.y > l2.s.y) swap(l1,l2);        int flag = l1.segcrossseg(l2);        if(flag == 0 || l1.parallel(l2) || sgn(l1.e.y - l1.s.y) == 0 || sgn(l2.e.y - l2.s.y) == 0)//与x平行，或无交点，或重合        {            printf("0.00\n"); continue;        }        Point jiao = l1.crosspoint(l2);        Point u = l1.s;        Point v = u; v.x += 1;        Line l3 = Line(u,v);        v = l2.crosspoint(l3);                if(sgn(l1.s.x - l1.e.x) != 0 && sgn(l2.s.x - l2.e.x) != 0)//完全覆盖        {            double k1 = l1.getk(),k2 = l2.getk();            if(k1*k2 > 0)            {                if(k1 > k2 && l1.s.x > l2.s.x - eps) goto stop;                if(k1 < k2 && l2.s.x > l1.s.x - eps) goto stop;            }        }        ans = (fabs(u.x-v.x)*fabs(u.y - jiao.y))/2;        printf("%.2f\n",ans); continue;        stop:printf("0.00\n");    }}/*106259 2664 8292 9080 1244 2972 9097 96800 1 1 01 0 2 10 1 2 11 0 1 20 0 10 100 0 9 80 0 10 100 0 8 90.9 3.1 4 00 3 2 20 0 0 20 0 -3 21 1 1 40 0 2 31 2 1 40 0 2 31 4 3 11 3 4 1输出：6162.651.000.000.004.500.503.000.750.000.00*/