题意：有两只青蛙和若干块石头，现在已知这些东西的坐标，两只青蛙A坐标和青蛙B坐标是第一个和第二个坐标，现在A青蛙想要到B青蛙那里去，并且A青蛙可以借助任意石头的跳跃，而从A到B有若干通路，问从A到B的所有通路上的最大边
链接：点我
floyd变形即可
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
using namespace std;
#define MOD 1000000007
#define pb(a) push_back(a)
const int INF=0x3f3f3f3f;
const double eps=1e-;
typedef long long ll;
#define cl(a) memset(a,0,sizeof(a))
#define ts printf("*****\n");
const int MAXN=;
int n,m,tt,cnt;
struct Node
{
int x,y;
void in()
{
scanf("%d%d",&x,&y);
}
}node[MAXN];
double dis(Node a,Node b)
{
return sqrt((double)(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double dist[MAXN][MAXN]; int main()
{
int i,j,k;
#ifndef ONLINE_JUDGE
freopen("1.in","r",stdin);
#endif
int ca=;
while(scanf("%d",&n)!=EOF)
{
if(n==) break;
for(i=;i<n;i++)
{
node[i].in();
}
cl(dist);
for(i=;i<n;i++)
{
for(j=;j<n;j++)
{
if(i==j) continue;
else dist[i][j]=dist[j][i]=dis(node[i],node[j]);
}
}
for(k=;k<n;k++)
{
for(i=;i<n;i++)
{
for(j=;j<n;j++)
{
if(dist[i][j]>max(dist[i][k],dist[k][j])) dist[i][j]=max(dist[i][k],dist[k][j]);
}
}
}
printf("Scenario #%d\nFrog Distance = %.3lf\n\n",ca++,dist[][]);
}
return ;
}