最短路(Floyd_Warshall) POJ 2253 Frogger

 

题目传送门

 1 /*
 2     最短路：Floyd算法模板题
 3 */
 4 #include <cstdio>
 5 #include <iostream>
 6 #include <algorithm>
 7 #include <cmath>
 8 #include <cstring>
 9 #include <string>
10 #include <vector>
11 using namespace std;
12 
13 const int MAXN = 200 + 10;
14 const int INF = 0x3f3f3f3f;
15 double d[MAXN][MAXN];
16 int used[MAXN];
17 struct NODE
18 {
19     int x, y;
20 }node[MAXN];
21 
22 double dis(int j, int i)
23 {
24     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));
25 }
26 
27 void Floyd_Warshall(int n)
28 {
29     for (int k=1; k<=n; ++k)
30     {
31         for (int i=1; i<=n-1; ++i)
32         {
33             for (int j=i+1; j<=n; ++j)
34             {
35                 if (d[i][k] < d[i][j] && d[k][j] < d[i][j])
36                 {
37                     if (d[i][k] < d[k][j])
38                         d[i][j] = d[j][i] = d[k][j];
39                     else
40                         d[i][j] = d[j][i] = d[i][k];
41                 }
42             }
43         }
44     }
45 
46     printf ("Frog Distance = %.3f\n", d[1][2]);
47 }
48 
49 int main(void)      //POJ 2253  Frogger
50 {
51     //freopen ("D.in", "r", stdin);
52 
53     int n;
54     int cnt = 0;
55     int first = 1;
56     while (~scanf ("%d", &n) && n)
57     {
58         for (int i=1; i<=n; ++i)
59         {
60             scanf ("%d%d", &node[i].x, &node[i].y);
61         }
62         for (int i=1; i<=n-1; ++i)
63         {
64             for (int j=i+1; j<=n; ++j)
65             {
66                 d[i][j] = d[j][i] = dis (i, j);
67             }
68         }
69 
70         printf ("Scenario #%d\n", ++cnt);
71         Floyd_Warshall (n);
72         puts ("");
73     }
74 
75     return 0;
76 }
77 
78 
79 
80 
81 /*
82 Scenario #1
83 Frog Distance = 5.000
84 
85 Scenario #2
86 Frog Distance = 1.414
87 */

 

posted @ 2015-03-27 18:26  Running_Time  阅读(147)  评论(0)  编辑  收藏  举报