poj2826 An Easy Problem?! 2012-01-11

http://acm.pku.edu.cn/JudgeOnline/problem?id=2826

 

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多种情况判断。如果用pascal最后输出答案时最好加个esp，即精度。不然有可能输出-0.00。

觉得算是比较标准的判断线段相交及求出交点的模板了。虽然有点冗长。

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Program Stone;
const exp=1e-8;
type coord=record
             x,y:real;
           end;
var n,i:longint;
    o0,o1,o2,o3,o4,o5,o6,o7,o8,oi,ox:coord;
    flag:boolean;
 Function cross(o1,o2,o3:coord):real;    //叉积
  begin
     cross:=(o2.x-o1.x)*(o3.y-o1.y)-(o3.x-o1.x)*(o2.y-o1.y);
  end;
 Function between(y1,y2,y3:real):boolean;
  Begin
    if (y1-y2)*(y1-y3)<exp then between:=true else between:=false;
  end;
 Function judgeintersect(o1,o2,o3,o4:coord):coord;    //判断相交和求出交点。
 var s1,s2,s3,s4:real;
  begin
     s1:=cross(o1,o2,o3);
     s2:=cross(o1,o2,o4);
     s3:=cross(o3,o4,o1);
     s4:=cross(o3,o4,o2);
     flag:=true;
     if (s1*s2<-exp)and(s3*s4<-exp) then
        begin
           judgeintersect.x:=((o3.x*s2)-(o4.x*s1))/(s2-s1);
           judgeintersect.y:=((o3.y*s2)-(o4.y*s1))/(s2-s1);  //运用规范相交公式求交点。
           exit;
        end;
     if (s1=0)and(between(o3.y,o1.y,o2.y)) then begin judgeintersect:=o3;exit;end;   //非规范相交，即交点为线段的某个端点。
     if (s2=0)and(between(o4.y,o1.y,o2.y)) then begin judgeintersect:=o4;exit;end;
     if (s3=0)and(between(o1.y,o3.y,o4.y)) then begin judgeintersect:=o1;exit;end;
     if (s4=0)and(between(o2.y,o3.y,o4.y)) then begin judgeintersect:=o2;exit;end;
     flag:=false;  //不然就不相交。
  end;
 Procedure work;
 var s:real;
     k:longint;
  begin
     k:=0;
     if (o1.y-oi.y>exp) then begin inc(k);o5:=o1;end;  
     if (o2.y-oi.y>exp) then begin inc(k);o5:=o2;end;
     if (o3.y-oi.y>exp) then begin inc(k);o6:=o3;end;
     if (o4.y-oi.y>exp) then begin inc(k);o6:=o4;end;   
     if k<2 then begin writeln('0.00');exit;end;
     if o5.y-o6.y>exp then begin o7:=o5;o5:=o6;o6:=o7;end;
     o0.x:=oi.x;o0.y:=oi.y+100;
     if ((cross(oi,o0,o6)*cross(oi,o6,o5)>exp))and((abs(o6.x-oi.x)-abs(o5.x-oi.x))>=-exp) then
      begin writeln('0.00');exit;end;
     o7.x:=100000000;o7.y:=o5.y;
     o8.x:=-100000000;o8.y:=o5.y;
     ox:=judgeintersect(o7,o8,o6,oi);
     s:=0;
     o0.x:=0;o0.y:=0;
     s:=(cross(o0,o5,oi)+cross(o0,oi,ox)+cross(o0,ox,o5))*0.5;
     writeln(abs(s)+exp:0:2);  //要加上esp
  end;
Begin
 assign(input,'input1.in');reset(input);
 assign(output,'output.out');rewrite(output);
   readln(n);
   for i:=1 to n do
    begin
       readln(o1.x,o1.y,o2.x,o2.y,o3.x,o3.y,o4.x,o4.y);
       oi:=judgeintersect(o1,o2,o3,o4);   //求出交点。
       if (abs(o1.x-o2.x)<exp)and(abs(o1.y-o2.y)<exp) then flag:=false;
       if (abs(o3.x-o4.x)<exp)and(abs(o3.y-o4.y)<exp) then flag:=false;
       if flag=false then writeln('0.00')
                          else work;
    end;
 close(input);close(output);
end.