NOIP2001T4car的旅行计划

洛谷传送门

一看数据就是floyed（毕竟年代久远），然而建图不是那么好贱好建，只知道三个机场，需要判断斜边来求第4个机场坐标。

往后一些麻烦的建图。

最后floyed就好。

——代码

 #include <iostream>
 #include <cstdio>
 #include <cmath>

 using namespace std;

 int n, s, a, b;
 ][], y[][], T[];
 ][], ans;

 inline int square(int ax)
 {
     return ax * ax;
 }

 inline int dis(int ax, int ay, int bx, int by)
 {
     return square(ax - bx) + square(ay - by);
 }

 int main()
 {
     int i, j, k;
     scanf("%d", &n);
     while(n--)
     {
         scanf("%d %d %d %d", &s, &t, &a, &b);
         ; i <  * s; i++)
          ; j <  * s; j++)
           map[i][j] = ;
         ans = ;
         ; i < s; i++)
         {
             scanf(], &y[i][], &x[i][], &y[i][], &x[i][], &y[i][], &T[i]);
             ], y[i][], x[i][], y[i][]) > dis(x[i][], y[i][], x[i][], y[i][])
             && dis(x[i][], y[i][], x[i][], y[i][]) > dis(x[i][], y[i][], x[i][], y[i][]))
             {
                 x[i][] = x[i][] + x[i][] - x[i][];
                 y[i][] = y[i][] + y[i][] - y[i][];
             }
             ], y[i][], x[i][], y[i][]) > dis(x[i][], y[i][], x[i][], y[i][])
                  && dis(x[i][], y[i][], x[i][], y[i][]) > dis(x[i][], y[i][], x[i][], y[i][]))
                  {
                      x[i][] = x[i][] + x[i][] - x[i][];
                      y[i][] = y[i][] + y[i][] - y[i][];
                  }
             ], y[i][], x[i][], y[i][]) > dis(x[i][], y[i][], x[i][], y[i][])
                  && dis(x[i][], y[i][], x[i][], y[i][]) > dis(x[i][], y[i][], x[i][], y[i][]))
                  {
                      x[i][] = x[i][] + x[i][] - x[i][];
                      y[i][] = y[i][] + y[i][] - y[i][];
                   }
             map[ * i][ * i + ] = map[ * i + ][ * i] = sqrt(dis(x[i][], y[i][], x[i][], y[i][])) * T[i];
             map[ * i][ * i + ] = map[ * i + ][ * i] = sqrt(dis(x[i][], y[i][], x[i][], y[i][])) * T[i];
             map[ * i][ * i + ] = map[ * i + ][ * i] = sqrt(dis(x[i][], y[i][], x[i][], y[i][])) * T[i];
             map[ * i + ][ * i + ] = map[ * i + ][ * i + ] = sqrt(dis(x[i][], y[i][], x[i][], y[i][])) * T[i];
             map[ * i + ][ * i + ] = map[ * i + ][ * i + ] = sqrt(dis(x[i][], y[i][], x[i][], y[i][])) * T[i];
             map[ * i + ][ * i + ] = map[ * i + ][ * i + ] = sqrt(dis(x[i][], y[i][], x[i][], y[i][])) * T[i];
             ; j <  * i; j++)
               * i; k <  * i + ; k++)
               map[k][j] = map[j][k] = sqrt(dis(x[j / ][j %  + ], y[j / ][j %  + ], x[k / ][k %  + ], y[k / ][k %  + ])) * t;
         }
         ; i <  * s; i++) map[i][i] = ;
         ; k <  * s; k++)
          ; i <  * s; i++)
           ; j <  * s; j++)
            map[i][j] = min(map[i][j], map[i][k] + map[k][j]);
          * a - ; i <  * a; i++)
           * b - ; j <  * b; j++)
           ans = min(ans, map[i][j]);
         printf("%.1f\n", ans);
     }
     ;
 }