POJ 2826 an easy problem?! Good question

The main idea is that two pieces of wood to form a slot, ask how much rain tank can be installed, pay attention to the vertical fall of rain, thinking is very simple, is the classification of the discussion is a bit bad.
1. If the two segments do not intersect or are parallel, install 0;
2. There is a parallel x-axis, installed 0;
3. If the top cover below, install 0;
4. Other, cross-product to find area.

Directly on the code:

#include <iostream>#include <cmath>#include <stdio.h>Using namespaceSTD;Const Double eps=1e-8;struct point{Doublex;Doubley;};struct line{Point A;Point B;}l1,l2;Double ans;Cross product double Xmult (point p0, point P1, point p2) {return (P1. x-p0. x) * (P2. Y-p0. Y)-(P2. x-p0. x) * (P1. Y-p0. Y);} int dblcmp (double n) {if (Fabs (n) <eps) return0;Return n>0?1:-1;}//determine if they intersect, how to intersect int judge (line L1, line L2) {Double d1=dblcmp (max (L1. A. x, L1. b. x)-min (L2. A. x, L2. b. x));Double d2=dblcmp (Max (L2. A. x, L2. b. x)-min (L1. A. x, L1. b. x));Double d3=dblcmp (Max (L1. A. Y, L1. b. Y)-min (L2. A. Y, L2. b. Y));Double d4=dblcmp (Max (L2. A. Y, L2. b. Y)-min (L1. A. Y, L1. b. Y));Double d5=dblcmp (Xmult (L2. A, L1. A, L1. b));Double d6=dblcmp (Xmult (L2. b, L1. A, L1. b));Double d7=dblcmp (Xmult (L1. A, L2. A, L2. b));Double d8=dblcmp (Xmult (L1. b, L2. A, L2. b));if (d1>=0&&d2>=0&&d3>=0&&d4>=0) {if (d5*d6>0|| D7*d8>0) return0;//Do not intersectelse if (d5==0&&d6==0) return1;//Common line intersectionelse if (d5==0|| d6==0|| d7==0|| d8==0) return2;//Endpoint intersectionelse return3;//Specification intersect} return0;}//slope bool Getslope (line L, double &k) {double t=l. A. x-L. b. x;if (t==0) return False;K= (l. A. Y-L. b. Y)/T;return True;}//the intersection point of Point Getintersect (lines L1, line L2) {point P;Double A1 = L1. b. Y-L1. A. Y;Double B1 = L1. A. x-L1. b. x;Double C1 = (L1. b. x-L1. A. x) * L1. A. Y-(L1. b. Y-L1. A. Y) * L1. A. x;Double A2 = L2. b. Y-L2. A. Y;Double B2 = L2. A. x-L2. b. x;Double C2 = (L2. b. x-L2. A. x) * L2. A. Y-(L2. b. Y-L2. A. Y) * L2. A. x;P. x= (C2*B1-C1*B2)/(A1*B2-A2*B1);P. Y= (C1*A2-C2*A1)/(A1*B2-A2*B1);if (p. x==-0) p. x=0;Return P;}//For a, b two pointyThe point at which the coordinates are larger dot getbiggery;if (DBLCMP (a. Y-B. Y) >0) {Q. x= A. x;Q. Y= A. Y;} else {Q. x= b. x;Q. Y= b. Y;} return Q;}double Getarea (point P, point P1, point p2) {point q;Double A;if (dblcmp (P1. Y-p2. Y) >=0) {Q. Y= P2. Y;Q. x= P. x+ (P1. x-P. x) * (P2. Y-P. Y)/(P1. Y-P. Y);//ask for the coordinates of another pointA=fabs (Xmult (p, p2, q))/2;//cross-product to calculate the area} else {Q. Y= P1. Y;Q. x= P. x+ (P2. x-P. x) * (P1. Y-P. Y)/(P2. Y-P. Y);A=fabs (Xmult (P, p1, q))/2;} return a;}int Main () {int T;Cin>>t;while (t--) {point P1, p2, p;Cin>>l1. A. x>>l1. A. Y>>l1. b. x>>l1. b. Y;Cin>>l2. A. x>>l2. A. Y>>l2. b. x>>l2. b. Y;if (judge (L1,L2) <=1)//whether intersect {ans=0; cout<<0<<endl;}else{P1=getbiggery (L1. A, L1. b);//cout << "(" <<p1.x<< "<<p1.y<<") "<<endl;P2=getbiggery (L2. A, L2. b);//cout << "(" <<p2.x<< "<<p2.y<<") "<<endl;P=getintersect (L1, L2);//cout << "(" <<p.x<< "<<p.y<<") "<<endl;if (!DBLCMP (p. x-p1. x) &AMP;&AMP;!DBLCMP (P. Y-p1. Y) || !DBLCMP (P. x-p2. x) &AMP;&AMP;!DBLCMP (P. x-p2. Y)//Opening direction//{ans=0;//cout<<1<<endl;}else{Double K1, K2;BOOL F1, F2;F1 = getslope (L1, K1);F2 = Getslope (L2, K2);if (!dblcmp (k1) | |!dblcmp (k2))//{ans =0;//cout<<2<<endl;} When a line segment is parallel to the x-axiselse if (F1 && F2) {//If both lines are notyShaft Parallel if (dblcmp (K1*K2) >0{//when the slope symbol of the two segments is the same, int d1 = dblcmp (K1-K2);int d2 = dblcmp (K2);if (d1>0&&d2>0&AMP;&AMP;DBLCMP (P2. x-p1. x) *dblcmp (P2. x-P. x) <=0|| d1<0&&d2>0&AMP;&AMP;DBLCMP (P1. x-p2. x) *dblcmp (P1. x-P. x) <=0|| D1>0&&d2<0&AMP;&AMP;DBLCMP (P1. x-p2. x) *dblcmp (P1. x-P. x) <=0|| d1<0&&d2<0&AMP;&AMP;DBLCMP (P2. x-p1. x) *dblcmp (P2. x-P. x) <=0)//Coverage condition//{ans =0;//cout<<3<<endl;}else//{Ans=getarea (P,P1,P2);//cout<<4<<endl;}} else//{Ans=getarea (P,P1,P2);//cout<<5<<endl;}} else//{Ans=getarea (P,P1,P2);//cout<<6<<endl;}}} printf ("%.2lf\n", ans);} return0;}

POJ 2826 an easy problem?! Good question