poj 2826 An Easy Problem?! (线段相交问题终极版...并不easy)

An Easy Problem?!

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11066		Accepted: 1673

Description

It's raining outside. Farmer Johnson's bull Ben wants some rain to water his flowers. Ben nails two wooden boards on the wall of his barn. Shown in the pictures below, the two boards on the wall just look like two segments on the plane, as they have the same width. 

Your mission is to calculate how much rain these two boards can collect. 

Input

The first line contains the number of test cases. 
Each test case consists of 8 integers not exceeding 10,000 by absolute value, x₁, y₁, x₂, y₂, x₃, y₃, x₄, y₄. (x₁, y₁), (x₂, y₂) are the endpoints of one board, and (x₃, y₃), (x₄, y₄) are the endpoints of the other one. 

Output

For each test case output a single line containing a real number with precision up to two decimal places - the amount of rain collected. 

Sample Input

2
0 1 1 0
1 0 2 1

0 1 2 1
1 0 1 2

Sample Output

1.00
0.00

感动到cry。。。终于A掉了。
我的做法是这样的：
有三个重要的点，一个是两条线段的交点，设为crossp
第二个是末端点中较低的点（所以读入的时候adjust一下，线段方向维护成由低指向高）,设为lowp
第三个是由lowp做平行线，交到另一条直线（anotherline)得到点p3

雨是垂直下落。
一般的情况只要求这三个点组成的三角形的面积就行了（叉积或者h*c/2都可以） 两种写法都可以A
主要是一些误解情况的判断。

首先，不相交的话，肯定无解。
所以先判断下相交。

细分的话，如果有一条直线平行x轴，那么肯定无解。

接下来是，crossp正好和lowp重合，肯定无解。

然而如果用c*h/2的话，2,3两种情况可以不用单独考虑，因为会得出h==0,答案自然为0

接下来比较难想到的是，口被封住的情况。

虽然我想到了这种情况，但是判定条件想错了。

我误以为两个末端点在p3的同侧的话，口就会无解。

但如下图

这种情况却是有解的，答案不为0

这步判定想错是我WA了多次最主要的原因QAQ

不过只要稍微改动下就好

只有x在lowp那条线的左边且lowp在anotherline末端点的左边

或者x在lowp那条线的右边且lowp在anotherline末端点的右边  口才会被封住。 

    int sameside(point p1,point p2)
    {
//    p1.output();
//    p2.output();
    if (dblcmp(x-p2.x)<=0&&dblcmp(p2.x-p1.x)<=0) return 1;
    if (dblcmp(x-p2.x)>=0&&dblcmp(p2.x-p1.x)>=0) return 1;
    return 0;
    }

因为一个细节又WA了一次。
上面的代码一开始忘写了等号。
如果但实际上末端点的横坐标相等，也是无解的。

/*************************************************************************
    > File Name: code/poj/2826.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年11月06日 星期五 16时07分21秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#include<cctype>
#define fst first              
#define sec second      
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
using namespace std;
const double eps = 1E-10;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
typedef long long LL;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
int dblcmp(double d)
{
    return d < -eps ? -1 : d > eps;
}
inline double sqr(double x){return x*x;}
struct point
{
    double x,y;
    point(){}
    point(double _x,double _y):
    x(_x),y(_y){};
    void input()
    {
        scanf("%lf%lf",&x,&y);
    }
    void output()
    {
        printf("%.2f %.2f
",x,y);
    }
    bool operator==(point a)const
    {
        return dblcmp(a.x-x)==0&&dblcmp(a.y-y)==0;
    }
    bool operator<(point a)const
    {
        return dblcmp(a.x-x)==0?dblcmp(y-a.y)<0:x<a.x;
    }
    double len()
    {
        return hypot(x,y);
    }
    double len2()
    {
        return x*x+y*y;
    }
    double distance(point p)
    {
        return hypot(x-p.x,y-p.y);
    }
    double distance2(point p)
    {
        return sqr(x-p.x)+sqr(y-p.y);
    }
    point add(point p)
    {
        return point(x+p.x,y+p.y);
    }
    point operator + (const point & p) const{
        return point(x+p.x,y+p.y);
    }
    point sub(point p)
    {
        return point(x-p.x,y-p.y);
    }
    point operator - (const point & p) const{
        return point(x-p.x,y-p.y);
    }
    point mul(double b)
    {
        return point(x*b,y*b);
    }
    point div(double b)
    {
        return point(x/b,y/b);
    }
    double dot(point p)
    {
        return x*p.x+y*p.y;
    }
    double operator * (const point & p) const{
        return x*p.x+y*p.y;
    }
    double det(point p)
    {
        return x*p.y-y*p.x;
    }
    double operator ^ (const point & p) const{
        return x*p.y-y*p.x;
    }
    double rad(point a,point b)
    {
        point p=*this;
        return fabs(atan2(fabs(a.sub(p).det(b.sub(p))),a.sub(p).dot(b.sub(p))));
    }
    point trunc(double r)
    {
        double l=len();
        if (!dblcmp(l))return *this;
        r/=l;
        return point(x*r,y*r);
    }
    point rotleft()
    {
        return point(-y,x);
    }
    point rotright()
    {
        return point(y,-x);
    }
    point rotate(point p,double angle)//绕点p逆时针旋转angle角度
    {
        point v=this->sub(p);
        double c=cos(angle),s=sin(angle);
        return point(p.x+v.x*c-v.y*s,p.y+v.x*s+v.y*c);
    }

    int sameside(point p1,point p2)
    {
//    p1.output();
//    p2.output();
    if (dblcmp(x-p2.x)<=0&&dblcmp(p2.x-p1.x)<=0) return 1;
    if (dblcmp(x-p2.x)>=0&&dblcmp(p2.x-p1.x)>=0) return 1;
    return 0;
    }
    double gettrianglearea(point b,point c)
    {
    return ((b.x-x)*(c.y-y)-(b.y-y)*(c.x-x))/2;
    }
};
struct line
{
    point a,b;
    line(){}
    line(point _a,point _b)
    {
        a=_a;
        b=_b;
    }
    bool operator==(line v)
    {
        return (a==v.a)&&(b==v.b);
    }
    //倾斜角angle
    line(point p,double angle)
    {
        a=p;
        if (dblcmp(angle-pi/2)==0)
        {
            b=a.add(point(0,1));
        }
        else
        {
            b=a.add(point(1,tan(angle)));
        }
    }
    //ax+by+c=0
    line(double _a,double _b,double _c)
    {
        if (dblcmp(_a)==0)
        {
            a=point(0,-_c/_b);
            b=point(1,-_c/_b);
        }
        else if (dblcmp(_b)==0)
        {
            a=point(-_c/_a,0);
            b=point(-_c/_a,1);
        }
        else
        {
            a=point(0,-_c/_b);
            b=point(1,(-_c-_a)/_b);
        }
    }
    void input()
    {
        a.input();
        b.input();
    }
    void adjust()  //调整为由低指向高
    {
    if (dblcmp(a.y-b.y)>0) swap(a,b);
    }
    double length()
    {
        return a.distance(b);
    }
    double getk()
    {
//    printf("*********
a.x: %f a.y %f b.x : %f b.y : %f
 ****************
",a.x,a.y,b.x,b.y);
    return (b.y-a.y)/(b.x-a.x);
    }
    double angle()//直线倾斜角 0<=angle<180
    {
        double k=atan2(b.y-a.y,b.x-a.x);
        if (dblcmp(k)<0)k+=pi;
        if (dblcmp(k-pi)==0)k-=pi;
//        cout<<"ang1:"<<k<<endl;
//        cout<<"k:"<<tan(k)<<endl;
        return k;
    }
    //点和线段关系
    //1 在逆时针
    //2 在顺时针
    //3 平行
    int relation(point p)
    {
        int c=dblcmp(p.sub(a).det(b.sub(a)));
        if (c<0)return 1;
        if (c>0)return 2;
        return 3;
    }
    bool pointonseg(point p)
    {
        return dblcmp(p.sub(a).det(b.sub(a)))==0&&dblcmp(p.sub(a).dot(p.sub(b)))<=0;
    }
    bool parallel(line v)
    {
        return dblcmp(b.sub(a).det(v.b.sub(v.a)))==0;
    }
    //2 规范相交
    //1 非规范相交 ()
    //0 不相交
    int segcrossseg(line v)
    {
        int d1=dblcmp(b.sub(a).det(v.a.sub(a)));
        int d2=dblcmp(b.sub(a).det(v.b.sub(a)));
        int d3=dblcmp(v.b.sub(v.a).det(a.sub(v.a)));
        int d4=dblcmp(v.b.sub(v.a).det(b.sub(v.a)));
        if ((d1^d2)==-2&&(d3^d4)==-2)return 2;
        return (d1==0&&dblcmp(v.a.sub(a).dot(v.a.sub(b)))<=0||
                d2==0&&dblcmp(v.b.sub(a).dot(v.b.sub(b)))<=0||
                d3==0&&dblcmp(a.sub(v.a).dot(a.sub(v.b)))<=0||
                d4==0&&dblcmp(b.sub(v.a).dot(b.sub(v.b)))<=0);
    }
    int linecrossseg(line v)//*this seg v line
    {
        int d1=dblcmp(b.sub(a).det(v.a.sub(a)));
        int d2=dblcmp(b.sub(a).det(v.b.sub(a)));
        if ((d1^d2)==-2)return 2;
        return (d1==0||d2==0);
    }
    point crosspoint(line v)
    {
        double a1=v.b.sub(v.a).det(a.sub(v.a));
        double a2=v.b.sub(v.a).det(b.sub(v.a));
        return point((a.x*a2-b.x*a1)/(a2-a1),(a.y*a2-b.y*a1)/(a2-a1));
    }
}li[10];
int main()
{
  #ifndef  ONLINE_JUDGE 
   freopen("in.txt","r",stdin);
  #endif

   int T;
   scanf("%d",&T);
   while (T--)
   {                                          
    li[1].input();li[1].adjust();
    li[2].input();li[2].adjust();
    if (li[1].segcrossseg(li[2])==0)  //不相交
    {
        puts("0.00");
        continue;
    }
    
    point crossp=li[1].crosspoint(li[2]);
    point lowp;
    int anotherline;
    if (li[1].b.y<li[2].b.y)
    {
        lowp=li[1].b;
        anotherline=2;
    }
    else
    {
        lowp=li[2].b;
        anotherline=1 ;
    }

    point p3;
    if (dblcmp(li[anotherline].a.x-li[anotherline].b.x)==0)
    {
        p3.x=li[anotherline].a.x;
        p3.y=lowp.y;
    }
    else
    {
        double dy =lowp.y-li[anotherline].a.y;
        double k = li[anotherline].getk();
        p3.x=li[anotherline].a.x+dy*1.0/k;
        p3.y=lowp.y;
    }
    double h = lowp.y-crossp.y;
    if (dblcmp(h)==0)
    {
        puts("0.00");
        continue;
    }
    if (p3.sameside(li[anotherline].b,li[anotherline%2+1].b)==1)
    {
        puts("0.00");
        continue;
    }
    double c = fabs(lowp.x-p3.x);
    
    double ans = h*c/2;
//    double ans = lowp.gettrianglearea(crossp,p3);
    printf("%.2f
",fabs(ans));


   }
  
   
 #ifndef ONLINE_JUDGE  
  #endif
  fclose(stdin);
    return 0;
}