poj 2253 Frogger （dijkstra最短路）

题目链接：http://poj.org/problem?id=2253

Frogger

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 25773		Accepted: 8374

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：有两只青蛙和若干块石头，其中一只青蛙想去拜访另一只青蛙，现在已知这些东西的坐标，两只青蛙坐标分别是第一个和第二个坐标，并且这只青蛙可以借助任意石头的跳跃，而两只青蛙之间有若干通路，问两只的所有通路上的最大边，然后在通过这些最大边来找最短路。

从起点到终点会有很多路径，每条路径上的边有一个最大值，求这些最大值中的最小值。

也就是更新的边要保持最大边。

特别注意：这里我也不是很懂的地方，在最后输出的时候用%.3lf就是死命的wa，而改成%.3f就轻松ac了，其实不过在poj上是可以过得了，如果过不了再改也是可以的了，我是在专题里面一直wa，表示很无奈~

 

不多说了，看代码~

 #include <stdio.h>
 #include <math.h>
 const int INF=;
 double map[][],node[],Min;
 int n,vis[];

 double Max(double a,double b)
 {
     return a>b?a:b;
 }

 void dijkstra()
 {
     int i,j,k,m;
     for (i=; i<n; i++)
     {
         node[i]=map[][i];
         vis[i]=;
     }
     vis[]=;
     for (k=; k<n; k++)
     {
         Min=INF;
         m=-;
         for (i=; i<n; i++)
             if (!vis[i])
             {
                 if (Min>node[i])
                 {
                     Min=node[i];
                     m=i;
                 }
             }
         if (m==-)
             break;
         vis[m]=;
         //tm=m;
         for (i=; i<n; i++)
         {
             if (!vis[i]&&Max(node[m],map[m][i])<node[i])
                 node[i]=Max(node[m],map[m][i]);

         }
     }
 }

 int main ()
 {
     double a[],b[];
     int count=,i,j;
     while (scanf("%d",&n),n)
     {
         for (i=; i<n; i++)
         {
             scanf("%lf%lf",&a[i],&b[i]);
         }
         for (i=; i<n; i++)
         {
             for (j=; j<n; j++)
             {
                 map[i][j]=map[j][i]=(a[i]-a[j])*(a[i]-a[j])+(b[i]-b[j])*(b[i]-b[j]);

             }
         }
         dijkstra();
         printf("Scenario #%d\n", ++count);
         printf("Frog Distance = %.3f\n\n", sqrt(node[]));
     }
     return ;
 }