CF1079D Barcelonian Distance

思路：

模拟。

实现：

 #include <bits/stdc++.h>
 using namespace std;
 const long long INF = ;
 double dis(double x1, double y1, double x2, double y2)
 {
     return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
 }
 int main()
 {
     int a, b, c;
     double x1, x2, y1, y2;
     while (cin >> a >> b >> c)
     {
         cin >> x1 >> y1 >> x2 >> y2;
         if (a ==  || b == )
         {
             printf("%.10f\n", fabs(x1 - x2) + fabs(y1 - y2));
             continue;
         }
         double n1 = x1, m1 = y2; // y1--y2
         double ty = (-a * x1 - c) / b;
         double tx = (-b * y2 - c) / a;
         double ans = fabs(x1 - x2) + fabs(y1 - y2);
         if (ty >= min(y1, y2) && ty <= max(y1, y2) && tx >= min(x1, x2) && tx <= max(x1, x2))
         {
             double tmp = fabs(y1 - ty) + fabs(x2 - tx) + dis(x1, ty, tx, y2);
             ans = min(ans, tmp);
         }
         double ty2 = (-a * x2 - c) / b;
         if (ty >= min(y1, y2) && ty <= max(y1, y2) && ty2 >= min(y1, y2) && ty2 <= max(y1, y2))
         {
             double tmp = fabs(y1 - ty) + fabs(ty2 - y2) + dis(x1, ty, x2, ty2);
             ans = min(ans, tmp);
         }
         double n2 = x2, m2 = y1; // x1--x2
         tx = (-b * y1 - c) / a;
         ty = (-a * x2 - c) / b;
         if (ty >= min(y1, y2) && ty <= max(y1, y2) && tx >= min(x1, x2) && tx <= max(x1, x2))
         {
             double tmp = fabs(x1 - tx) + fabs(y2 - ty) + dis(tx, y1, x2, ty);
             ans = min(ans, tmp);
         }
         double tx2 = (-b * y2 - c) / a;
         if (tx >= min(x1, x2) && tx <= max(x1, x2) && tx2 >= min(x1, x2) && tx2 <= max(x1, x2))
         {
             double tmp = fabs(x1 - tx) + fabs(tx2 - x2) + dis(tx, y1, tx2, y2);
             ans = min(ans, tmp);
         }
         printf("%.10f\n", ans);
     }
     return ;
 }