这题也可以用Dijkstra算法做，我用prim做出来的，个人感觉prim更适合这道题。找到最小生成树里的最长边即可。注意并不是所有点都需要建成最小生成树，到达终点就可以终止了。我习惯用邻接表，而且顶点是从0开始的。

贴代码

//Memory 924k Time  32ms
#include<iostream>
#include<vector>
#include<queue>
#include<cmath>
using namespace std;

class road
{
public:
	int end;
	double weight;
};
class Graph
{
public:
	int v;             //顶点个数
	int mark[210];          //标记是否访问过，访问过为1，否则为0
	vector<road> *adj;  //邻接表

	Graph(int n);          //构造函数
	~Graph();                           //析构函数
	void addEdge(int b,int e,double w);    //添加边
};
Graph::Graph(int n)
{
	v=n; 
	adj=new vector<road>[n];
}
Graph::~Graph()
{
	delete []adj;
}
void Graph::addEdge(int b,int e,double w)    //添加边
{
	road r1;
	r1.end = e;
	r1.weight = w;
	adj[b].push_back(r1);

}

class Dist
{
public:
	int pre;
	double length;
	int index;
	friend bool operator<(const Dist &a,const Dist &b)
	{
		return a.length>b.length;
	}
};

double prim(Graph &g,int s)    //s是起始点
{
	double max_l=-1<<30;
	Dist * D = new Dist[g.v];   //申请空间
	for(int i=0;i<g.v;i++)        //初始化
	{
		D[i].index  = i;
		D[i].length = 1<<30;
		D[i].pre = s;
	}

	priority_queue<Dist> aqueue;
	int now_node = s;
	for(int i=0;i<g.v-1;i++)
	{
		vector<road>::iterator ii = g.adj[now_node].begin();
		for(;ii!=g.adj[now_node].end();ii++)    //刷新与now_node相邻的顶点的值
		{
			if( g.mark[ii->end]==0 && D[ii->end].length > ii->weight )
			{
				D[ii->end].length = ii->weight;
				D[ii->end].pre = now_node;
				aqueue.push(D[ii->end]);
			}
		}
		Dist d;
		bool FOUND=false;
		while(!aqueue.empty())
		{
			d = aqueue.top();
			aqueue.pop();
			if( g.mark[d.index]== 0)
			{
				break;
			}
		}

		g.mark[now_node]=1;
		now_node=d.index;
	
		max_l = max(max_l,d.length);
		if(now_node==1)          //到达终点
			break;
	}
	return max_l;
}

int main()
{
	int n;
	int num=1;
	int x[210];
	int y[210];
	double temp;
	while(cin>>n)
	{
		if(n==0)
			break;
		Graph g(n);
		memset(g.mark,0,sizeof(g.mark));
		memset(x,0,sizeof(x));
		memset(y,0,sizeof(y));
		for(int i=0;i<n;i++)
		{
			cin>>x[i]>>y[i];
		}

		for(int i=0;i<n;i++)
		{
			for(int j=0;j<n;j++)
			{
				if(i==j)
					continue;
				temp = sqrt( (double)(x[i]-x[j])*(x[i]-x[j]) +(double)(y[i]-y[j])*(y[i]-y[j]) );
				g.addEdge(i,j,temp);
			}
		}
		temp=prim(g,0);
		cout<<"Scenario #"<<num<<endl;
		printf("Frog Distance = %.3lf\n",temp);;
		cout<<endl;
		num++;
	}
	cout<<endl;
	return 0;
}