An Easy Problem?! poj 2826 一道全是坑点的题！！！

http://poj.org/problem?id=2826

An Easy Problem?!

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12059 Accepted: 1836

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

20 1 1 01 0 2 10 1 2 11 0 1 2

Sample Output

1.000.00

对了之后怒交三发才缓过来 (╯‵□′)╯︵┻━┻ 谨记自己是弱鸡；

题意 给两条线 这两条线组成的图形来兜住雨水 求雨水的面积

坑点：

兜不到 上边的x >= 下边的x 同理右边小于

还有 平行，共线 ，没有交点

还可以看 discuss 也有很多值得注意的

下边是几个样例

1000
0 1 1 0
1 0 2 1

0 1 2 1
1 0 1 2

0 0 1 1
1 1 2 0

6259 2664 8292 9080 1244 2972 9097 9680

0 1 1 0
1 0 2 1

0 1 2 1
1 0 1 2

0 0 10 10
0 0 9 8

0 0 10 10
0 0 8 9

0.9 3.1 4 0
0 3 2 2

0 0 0 2
0 0 -3 2

1 1 1 4
0 0 2 3

1 2 1 4
0 0 2 3

68.22 191.66 257.23 104.97
140.25 33.53 56.71 302.58

10 10 0 0 
0 0 1 1 

1 1 0 0
1 1 2 2

1 1 0 0
2 2 3 3

这是运行结果

1000
0 1 1 0
1 0 2 1
1.00

0 1 2 1
1 0 1 2
0.00

0 0 1 1
1 1 2 0
0.00

6259 2664 8292 9080 1244 2972 9097 9680
6162.65

0 1 1 0
1 0 2 1
1.00

0 1 2 1
1 0 1 2
0.00

0 0 10 10
0 0 9 8
0.00

0 0 10 10
0 0 8 9
4.50

0.9 3.1 4 0
0 3 2 2
0.50

0 0 0 2
0 0 -3 2
3.00

1 1 1 4
0 0 2 3
0.75

1 2 1 4
0 0 2 3
0.00

68.22 191.66 257.23 104.97
140.25 33.53 56.71 302.58
0.00

10 10 0 0
0 0 1 1
0.00

1 1 0 0
1 1 2 2
0.00

1 1 0 0
2 2 3 3
0.00 

//http://blog.sina.com.cn/s/blog_71dbfe2e0101f7zb.html//http://dev.gameres.com/Program/Abstract/Geometry.htm#矢量的概念#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>#include<stdlib.h>#include<math.h>using namespace std;const double eps=1e-8;#define zero(x) (((x) > 0 ? (x) : (-x)) < eps)double max(double a,double b){if(a > b) return a;return b ; }double min(double a,double b){if(a < b) return a;return b ; }struct point2d//点也是向量 {double x,y;point2d(double x=0,double y=0):x(x),y(y){}};struct polygon//多边形 点的集合; {point2d pp[1010];int size; };int isbanana(point2d p1,point2d p2,point2d p3,point2d p4);int fun(point2d a,point2d b,point2d c,point2d d);  int isclw(point2d a,point2d b,point2d c);point2d  line_cross(point2d a,point2d b,point2d c,point2d d);polygon p_h(polygon po,point2d a,point2d b);polygon p_hop(polygon po,point2d a,point2d b);void see(polygon pol);int  ispolcer(polygon pol);double dis(point2d a,point2d b);void see_p(point2d a);//重载point2d operator + (point2d a,point2d b){return point2d(a.x+b.x,a.y+b.y); }point2d operator - (point2d a,point2d b){return point2d(a.x-b.x,a.y-b.y); }point2d operator * (point2d a,double b){return point2d(a.x*b,a.y*b); }point2d operator / (point2d a,double b){return point2d(a.x/b,a.y/b); }//×乘 //矢量叉积：////　　计算矢量叉积是与直线和线段相关算法的核心部分。//设矢量P = ( x1, y1 )，Q = ( x2, y2 )，则矢量叉积定义为由(0,0)、//p1、p2和p1+p2所组成的平行四边形的带符号的面积，即：P × Q = x1*y2 - x2*y1，//其结果是一个标量。显然有性质 P × Q = - ( Q × P ) 和 P × ( - Q ) = - ( P × Q )。//一般在不加说明的情况下，本文下述算法中所有的点都看作矢量，//两点的加减法就是矢量相加减，而点的乘法则看作矢量叉积。//　　叉积的一个非常重要性质是可以通过它的符号判断两矢量相互之间的顺逆时针关系：//　　若 P × Q > 0 , 则P在Q的顺时针方向。//　　若 P × Q < 0 , 则P在Q的逆时针方向。//　　若 P × Q = 0 , 则P与Q共线，但可能同向也可能反向。double operator * (point2d a,point2d b){return a.x*b.y-b.x*a.y; }//·乘double pdp(point2d a,point2d b){return a.x*b.x+a.y*b.y; }  double dis(point2d a,point2d b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); }  //两线段是否平行int isl_l(point2d aa,point2d bb,point2d cc,point2d dd){point2d ab = aa-bb;point2d cd = cc-dd;if(cd*ab){return 0;}return 1; } //int main()//{////point2d a,b,c,d;//while(~scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y))//{//scanf("%lf%lf%lf%lf",&c.x,&c.y,&d.x,&d.y);//printf("is %d\n",isl_l(a,b,c,d));////}//}//两线段是否香蕉(包括边界香蕉)，快速排斥&&跨立实验 int isbanana(point2d aa,point2d bb,point2d cc,point2d dd) {int f1,f2;    f1=fun(aa,bb,cc,dd);      f2=fun(cc,dd,aa,bb);      if((f1==1&&f2==1)||f1==0||f2==0||(f1==0&&f2==0))         return 1;    else         return 0;}int fun(point2d a,point2d b,point2d c,point2d d)  {      point2d p[3];      int flag,m,n;      p[0].x=b.x-a.x;      p[0].y=b.y-a.y;      p[1].x=c.x-a.x;      p[1].y=c.y-a.y;      p[2].x=d.x-a.x;      p[2].y=d.y-a.y;      m=(p[0].x*p[1].y-p[0].y*p[1].x);      n=(p[0].x*p[2].y-p[0].y*p[2].x);      if((m%1000)!=0)          m=m%1000;      else      {          while((fabs((double)m/10))>10)              m=m/10;      }      if((n%1000)!=0)          n=n%1000;      else      {          while((fabs((double)n/10))>10)              n=n/10;      }      if(n*m<0)          flag=1;      else if(m*n==0)      {          if((c.x>=min(a.x,b.x)&&c.x<=max(a.x,b.x))&&(c.y>=min(a.y,b.y)&&c.y<=max(a.y,b.y))||(d.x>=min(a.x,b.x)&&d.x<=max(a.x,b.x))&&(d.y>=min(a.y,b.y)&&d.y<=max(a.y,b.y)))              flag=0;          else              flag=-1;      }      else          flag=-1;      return flag;  }   //clockwise;也可以判断a b c是否在同一直线或c在ab那一侧// ab有顺序 int isclw(point2d a,point2d b,point2d c){double ans = (a-b)*(b-c);if(ans<0)//叉乘< 0 为顺时针返回1  return 1;if(ans==0)//同一直线   return 0;if(ans>0) return -1;  } //int main()//{//point2d a,b,c;//a = point2d(0,0);//b = point2d(1,0);//c = point2d(1,-1);//printf("%d\n",isclw(a,b,c));//}//两直线焦点 point2d  line_cross(point2d a,point2d b,point2d c,point2d d){double a1,b1,c1,a2,b2,c2;//a1 = a.y - b.y;//b1 = b.x - a.x;//c1 = a.x*b.y - b.x*a.y;  //a*b;//a2 = c.y - d.y;//b2 = d.x - c.x;//c1 = c.x*d.y - d.x*c.y;//c*d; //double dd = a1 * b2 - a2 * b1;//printf("dd %f\n",dd);//return point2d((b1*c2-b2*c1)/dd,(c1*a2-c2*a1)/dd);//a1 = (a.y - b.y) / (a.x - b.x);//b1 = a.y - a1*(a.x);//a2 = (c.y - d.y) / (c.x - d.x);//b2 = c.y - a1*(c.x);//double xx = (b1 - b2) / (a2 - a1);//double yy = a1*xx+b1;//return point2d(xx,yy);//double lamuda = (fabs((d-a)*(b-a)))/(fabs((c-a)*(b-a)));//double xx = c.x+lamuda*(d.x-c.x);//double yy = c.y+lamuda*(d.y-c.y);// return point2d(xx,yy); double s2 = fabs((double)((d-a)*(b-a)))/2; double s1 = fabs((double)((c-a)*(b-a)))/2; return point2d((d.x*s1+c.x*s2)/(s1+s2) ,(d.y*s1+c.y*s2)/(s1+s2));  }//一个多边形与一个半平面的交集 polygon & halfpolygon p_h(polygon po,point2d a,point2d b){//printf("po.size %d\n",po.size); polygon ans; ans.size = 0; for(int i = 0;i < po.size;i++) { if(isclw(a,b,po.pp[i])>=0) { ans.pp[ans.size++] = po.pp[i]; }else if(isclw(a,b,po.pp[i])<0&&isclw(a,b,po.pp[(i+1)%po.size])>0){ans.pp[ans.size++] = line_cross(a,b,po.pp[i],po.pp[(i+1)%po.size]); }  }// printf("ans %d\n",ans.size); return ans;}void see(polygon pol){printf("size %d\n",pol.size);for(int i = 0;i < pol.size;i++) { printf("%.2f %.2f \n",pol.pp[i].x,pol.pp[i].y); }}void see_p(point2d a){printf("\n point %.2f %.2f\n",a.x,a.y);} //多边形有无核 int  ispolcer(polygon pol){polygon now = pol;point2d a,b;for(int i = 0;i < pol.size;i++){now = p_h(now,pol.pp[i],pol.pp[(i+1)%pol.size]);//printf("!ans %d",now.size);//see(now);if(now.size == 0) { return 0; }//printf("i %d\n",i);}return 1;  }    int main(){int T;scanf("%d",&T);while(T--){point2d a,b,c,d;scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);scanf("%lf%lf%lf%lf",&c.x,&c.y,&d.x,&d.y);if(isl_l(a,b,c,d)){printf("0.00\n");continue;}if(a.y==b.y||c.y==d.y){printf("0.00\n");continue;}if(!isbanana(a,b,c,d)){printf("0.00\n");continue;}point2d cross = line_cross(a,b,c,d);double abmaxy = max(a.y,b.y);double cdmaxy = max(c.y,d.y);point2d y1 = a;if(a.y<b.y) y1 = b;point2d y2 = c;if(c.y<d.y) y2 = d;point2d yy = y1;point2d yy_ = y2;if(y1.y<y2.y) {yy = y2; yy_ = y1;}if(cross.x<y1.x&&cross.x<y2.x){if(yy.x>=yy_.x&&yy.y>=yy_.y&&isclw(cross,yy_,yy)<0){printf("0.00\n");continue;} }if(cross.x>y1.x&&cross.x>y2.x&&isclw(cross,yy_,yy)>0){if(yy.x<=yy_.x&&yy.y>=yy_.y){printf("0.00\n");continue;} }double h = min(abmaxy,cdmaxy);point2d d1,d2;d1 = line_cross(a,b,point2d(-99999,h),point2d(99999,h));d2 = line_cross(c,d,point2d(-99999,h),point2d(99999,h));double dd = dis(d1,d2);printf("%.2f\n",(double)(h-cross.y)*dd/2);}}