(poj 2253) Frogger 最短路上的最大路段

题目链接：http://poj.org/problem?id=2253

Description
Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists‘ sunscreen, he wants to avoid swimming and instead reach her by jumping.
Unfortunately Fiona‘s stone is out of his jump range. Therefore Freddy considers to use other stones as intermediate stops and reach her by a sequence of several small jumps.
To execute a given sequence of jumps, a frog‘s jump range obviously must be at least as long as the longest jump occuring in the sequence.
The frog distance (humans also call it minimax distance) between two stones therefore is defined as the minimum necessary jump range over all possible paths between the two stones.

You are given the coordinates of Freddy‘s stone, Fiona‘s stone and all other stones in the lake. Your job is to compute the frog distance between Freddy‘s and Fiona‘s stone.

Input
The input will contain one or more test cases. The first line of each test case will contain the number of stones n (2<=n<=200). The next n lines each contain two integers xi,yi (0 <= xi,yi <= 1000) representing the coordinates of stone #i. Stone #1 is Freddy‘s stone, stone #2 is Fiona‘s stone, the other n-2 stones are unoccupied. There‘s a blank line following each test case. Input is terminated by a value of zero (0) for n.

Output
For each test case, print a line saying "Scenario #x" and a line saying "Frog Distance = y" where x is replaced by the test case number (they are numbered from 1) and y is replaced by the appropriate real number, printed to three decimals. Put a blank line after each test case, even after the last one.

Sample Input

2
0 0
3 4

3
17 4
19 4
18 5

0

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

Source
Ulm Local 1997

题目大意：有一只青蛙想到另一只青蛙那，但是不能一次到位，他需要借助中间的石头跳跃，问他到另一只那走的最短路中的最长跳跃距离是多少？

方法：求到另一只青蛙的地方的最小生成树的最大距离。

 1 #include<stdio.h>
 2 #include<cstring>
 3 #include<cstdlib>
 4 #include<algorithm>
 5 #include<math.h>
 6 #include<queue>
 7 using namespace std;
 8 #define INF 0x3f3f3f3f
 9 #define ll long long
10 #define met(a,b) memset(a,b,sizeof(a))
11 #define N 500
12 struct node
13 {
14     int x,y;
15
16 }s[N];
17 int Map[N][N];
18 int dis[N],ans;
19 int vis[N];
20 void dij(int n)
21 {
22     met(vis,0);
23     for(int i=1;i<=n;i++)
24         dis[i]=Map[1][i];
25     vis[1]=1;
26     for(int i=1;i<n;i++)
27     {
28         int an=INF;
29         int k=-1;
30         for(int j=1;j<=n;j++)
31         {
32             if(!vis[j] && an>dis[j])
33                 an=dis[k=j];
34         }
35         ans=max(ans,an);
36         if(k==2)///当k==2已经到达另一只青蛙的坐标
37             return ;
38             vis[k]=1;
39         for(int j=1;j<=n;j++)
40         {
41             if(!vis[j])
42                 dis[j]=min(dis[j],Map[k][j]);
43         }
44
45     }
46 }
47 int main()
48 {
49     int n;
50     int con=1;
51     while(scanf("%d",&n),n)
52     {
53         for(int i=0;i<=n;i++)
54         {
55             for(int j=0;j<=n;j++)
56                 Map[i][j]=i==j?0:INF;
57         }
58         for(int i=1;i<=n;i++)
59         {
60             scanf("%d %d",&s[i].x,&s[i].y);
61         }
62         for(int i=1;i<=n;i++)
63         {
64             for(int j=i;j<=n;j++)
65             {
66                 int ans=(s[i].x-s[j].x)*(s[i].x-s[j].x)+(s[i].y-s[j].y)*(s[i].y-s[j].y);
67                 Map[i][j]=Map[j][i]=ans;
68             }
69         }
70         ans=-INF;
71         dij(n);
72         printf("Scenario #%d\n",con++);
73         printf("Frog Distance = %.3f\n\n",sqrt(ans));
74     }
75     return 0;
76 }