最短路(Floyd_Warshall) POJ 2253 Frogger

题目传送门

 /*
     最短路：Floyd算法模板题
 */
 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cmath>
 #include <cstring>
 #include <string>
 #include <vector>
 using namespace std;

 const int MAXN =  + ;
 const int INF = 0x3f3f3f3f;
 double d[MAXN][MAXN];
 int used[MAXN];
 struct NODE
 {
     int x, y;
 }node[MAXN];

 double dis(int j, int i)
 {
     return sqrt ((node[j].x - node[i].x) * (node[j].x - node[i].x) + (node[j].y - node[i].y) * (node[j].y - node[i].y));
 }

 void Floyd_Warshall(int n)
 {
     for (int k=; k<=n; ++k)
     {
         for (int i=; i<=n-; ++i)
         {
             for (int j=i+; j<=n; ++j)
             {
                 if (d[i][k] < d[i][j] && d[k][j] < d[i][j])
                 {
                     if (d[i][k] < d[k][j])
                         d[i][j] = d[j][i] = d[k][j];
                     else
                         d[i][j] = d[j][i] = d[i][k];
                 }
             }
         }
     }

     printf ("Frog Distance = %.3f\n", d[][]);
 }

 int main(void)      //POJ 2253  Frogger
 {
     //freopen ("D.in", "r", stdin);

     int n;
     int cnt = ;
     int first = ;
     while (~scanf ("%d", &n) && n)
     {
         for (int i=; i<=n; ++i)
         {
             scanf ("%d%d", &node[i].x, &node[i].y);
         }
         for (int i=; i<=n-; ++i)
         {
             for (int j=i+; j<=n; ++j)
             {
                 d[i][j] = d[j][i] = dis (i, j);
             }
         }

         printf ("Scenario #%d\n", ++cnt);
         Floyd_Warshall (n);
         puts ("");
     }

     return ;
 }

 /*
 Scenario #1
 Frog Distance = 5.000

 Scenario #2
 Frog Distance = 1.414
 */