POJ2253 Frogger

解题思路：floyed模板题（稍微变换一点），不解释，上代码：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<cmath>
 5 using namespace std;
 6 #define inf 0x3f3f3f3f
 7 const int maxn = 205;
 8 double w[maxn][maxn];
 9
10 struct node{
11     double x, y;
12 }p[maxn];
13
14 int main()
15 {
16     int n, kase = 1;
17     while(~scanf("%d", &n) && n)
18     {
19         for(int i = 1; i <= n; i++)
20         for(int j = i; j <= n; j++)
21         {
22             if(i == j) w[i][j] = 0;
23             else w[i][j] = w[j][i] = inf;
24         }
25
26         for(int i = 1; i <= n; i++)
27         scanf("%lf %lf", &p[i].x, &p[i].y);
28
29         for(int i = 1; i <= n; i++)
30         {
31             for(int j = i; j <= n; j++)
32             {
33                 double x2 = p[j].x - p[i].x;
34                 double y2 = p[j].y - p[i].y;
35
36                 w[i][j] = w[j][i] = sqrt(x2*x2 + y2*y2);
37             }
38         }
39
40         for(int i = 1; i <= n; i++)
41         {
42             for(int j = 1; j <= n; j++)
43             {
44                 for(int k = j; k <= n; k++)
45                 {
46                     if(w[j][i] <= w[j][k] && w[i][k] <= w[j][k])
47                     {
48                         w[j][k] = w[k][j] = max(w[j][i], w[i][k]);
49                     }
50                 }
51             }
52         }
53
54         printf("Scenario #%d\nFrog Distance = %.3lf\n\n", kase++, w[1][2]);
55     }
56     return 0;
57 }