题目大致意思就是1想见2，它可以通过石头来中转，让路径中每次跳的最大距离最小。输出极限距离。
AC代码：

#include<iostream>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
const	int inf=0x3f3f3f3f;
const	int N=1010;
double d[N],w[N][N];
bool vis[N];
int n;
struct	node{
   
	double x;
	double y;
}now[N];
void dsj()
{
   
	memset(d,0x3f,sizeof(d));
	d[1]=0.0;
	for(int i=2;i<=n;i++)
	d[i]=w[1][i];
	vis[1]=1;
	for(int i=1;i<=n;i++)
	{
   
		int t=-1;
		for(int j=1;j<=n;j++)
		{
   
			if(!vis[j]&&(t==-1||d[t]>d[j]))
			t=j;
		}
			vis[t]=1;
			for(int j=1;j<=n;j++)
			{
   
				if(d[j]>max(d[t],w[t][j])&&!vis[j]&&w[t][j]!=inf)
				d[j]=max(d[t],w[t][j]);
			// printf("%lf",d[2]);
			}
}
}

int main()
{
   
	int k=0;
	while(scanf("%d",&n)!=EOF)
	{
   
		memset(vis,0,sizeof(vis));
		memset(w,0x3f,sizeof(w));
		if(n==0)	break;
		for(int i=1;i<=n;i++)
		{
   
			scanf("%lf%lf",&now[i].x,&now[i].y);
		}
		for(int i=1;i<=n;i++)
		{
   
			for(int j=1;j<=n;j++)
			w[i][j]=w[j][i]=sqrt((now[i].x-now[j].x)*(now[i].x-now[j].x)+(now[i].y-now[j].y)*(now[i].y-now[j].y));
		}
		dsj();
		printf("Scenario #%d\n",++k);
		printf("Frog Distance = %.3lf\n\n",d[2]);
	}
	return 0;
}

flyod做法

#include<algorithm>
#include<cmath>
#include<stdio.h>
#include<cstring>
#include<iostream>
using namespace std;
const	int N=1010;
double g[N][N];
int m,n;
double d[N];
bool st[N];
struct node{
   
	int x;
	int y;
}frog[N];
void floyd()
{
   
	for(int k=1;k<=n;k++)
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++)
	g[i][j]=min(g[i][j],max(g[i][k],g[k][j]));
}
int main()
{
   
	int k=0;
	while(cin>>n,n)
	{
   
		memset(st,0,sizeof st);
		memset(g,0,sizeof g);
		for(int i=1;i<=n;i++)
		{
   
			int x,y;
			cin>>x>>y;
			frog[i]={
   x,y};
		}
		for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
		g[j][i]=g[i][j]=sqrt(1.0*(frog[i].x-frog[j].x)*(frog[i].x-frog[j].x)+(frog[i].y-frog[j].y)*(frog[i].y-frog[j].y));
		floyd();
		printf("Scenario #%d\n",++k);
		printf("Frog Distance = %.3f\n\n",g[1][2]);
	}
	return 0;
 }