Codeforces 1079D Barcelonian Distance（计算几何）

题目链接：Barcelonian Distance

题意：给定方格坐标，方格坐标上有两个点A，B和一条直线。规定：直线上沿直线走，否则沿方格走。求A到B的最短距离。

题解：通过直线到达的：A、B两点都有两种方式到直线上，最多4种情况，每种情况求出A、B点到直线的距离和直线上新的两点间距离，取4种情况中最优的。

不通过直线到达：$abs(x1-x2)+abs(y1-y2)$，最后与通过直线到达的最优情况比较，得到最优解。

 #include <cmath>
 #include <cstdio>
 #include <algorithm>
 using namespace std;

 double a,b,c,ans;
 double x1,y1,x2,y2;

 double calx(double y){
     return (-c-b*y)/a;
 }

 double caly(double x){
     return (-c-a*x)/b;
 }

 double cal1(double x1,double x2){
     double X1,X2,Y1,Y2;
     double res=;
     X1=x1;Y1=caly(x1);
     res+=abs(Y1-y1);
     X2=x2;Y2=caly(x2);
     res+=abs(Y2-y2);
     res+=sqrt((X1-X2)*(X1-X2)+(Y1-Y2)*(Y1-Y2));
     return res;
 }

 double cal2(double x1,double y2){
     double X1,X2,Y1,Y2;
     double res=;
     X1=x1;Y1=caly(x1);
     res+=abs(Y1-y1);
     X2=calx(y2);Y2=y2;
     res+=abs(X2-x2);
     res+=sqrt((X1-X2)*(X1-X2)+(Y1-Y2)*(Y1-Y2));
     return res;
 }

 double cal3(double y1,double x2){
     double X1,X2,Y1,Y2;
     double res=;
     X1=calx(y1);Y1=y1;
     res+=abs(X1-x1);
     X2=x2;Y2=caly(x2);
     res+=abs(Y2-y2);
     res+=sqrt((X1-X2)*(X1-X2)+(Y1-Y2)*(Y1-Y2));
     return res;
 }

 double cal4(double y1,double y2){
     double X1,X2,Y1,Y2;
     double res=;
     X1=calx(y1);Y1=y1;
     res+=abs(X1-x1);
     X2=calx(y2);Y2=y2;
     res+=abs(X2-x2);
     res+=sqrt((X1-X2)*(X1-X2)+(Y1-Y2)*(Y1-Y2));
     return res;
 }

 int main(){
     scanf("%lf%lf%lf",&a,&b,&c);
     scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
     double ans=abs(x1-x2)+abs(y1-y2);
     if(a==||b==){
         printf("%.10f\n",abs(x1-x2)+abs(y1-y2));
         return ;
     }
     ans=min(ans,cal1(x1,x2));
     ans=min(ans,cal2(x1,y2));
     ans=min(ans,cal3(y1,x2));
     ans=min(ans,cal4(y1,y2));
     printf("%.10f\n",ans);
     return ;
 }