UVA 10803 Thunder Mountain

Original question:
Markus is building a army to fight the evil Valhalla Sector, so he needs to move some supplies between several of the NEA Rby towns. The woods is full of robbers and other unfriendly folk, so it's dangerous to travel far. As Thunder Mountain ' s head of security, Lee thinks that it's unsafe to carry supplies for more than 10km without visiting a town. Markus wants to know what far one would need-to-travel-get-from-one town-to-another in the worst case.
Input
The first line of input gives the number of cases, N. n test Cases follow. Each one starts with a line containing n (the number of towns, 1 < n < 101). The next n lines would give the xy-locations of each of the all the town's (integers in the range [0,1023]). Assume the Earth is flat and the whole 1024x1024 grid are covered by a forest with roads connecting each pair of tow NS that is no further than 10km away from each of the other.
Output
For each test case, output the line ' case #x: ', where x is the number of the The. On the next line, print the maximum distance one have-to-travel from town-to-town-B (for some A and b). Round the answer to 4 decimal places. Every answer would obey the formula |ans∗10 4−⌊ans∗10 4⌋−0.5| > 10−2
If It is a impossible to get from the some town to some, print ' Send kurdy ' instead. Put an empty line after each test case.
Sample Input
2
5
0 0
10 0
10 10
13 10
13 14
2
0 0
10 1
Sample Output
Case #1:
25.0000
Case #2:
Send Kurdy

English:
Give you a bunch of node coordinates, if the distance between two nodes is more than 10 then the two nodes are considered unreachable. Now ask you whether the given graph is connected, if not connected, output send Kurdy, otherwise calculate the shortest path of all nodes in this graph to other nodes, and then find the shortest path in the longest.

#include<bits/stdc++.h>
using namespace std;
struct point
{
    double x,y;
};
point points[110];
int n;
double G[110][110];
double dist(const point &a,const point &b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

void floyed()
{
    double ans=0;
    int ii,jj;
    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],G[i][k]+G[k][j]);
            }
        }
    }
}
int main()
{
    ios::sync_with_stdio(false);
    int t;
    cin>>t;
    int k=1;
    while(t--)
    {
        cin>>n;
        for(int i=1;i<=n;i++)
            cin>>points[i].x>>points[i].y;
        memset(G,0,sizeof(G));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                double dis=dist(points[i],points[j]);
                if(dis<=10)
                    G[i][j]=dis;
                else
                    G[i][j]=INT_MAX;
            }
        }
        floyed();
        double ans=-1;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
                ans=max(ans,G[i][j]);
        }
        if(ans>=INT_MAX)
        {
            cout<<"Case #"<<k++<<':'<<endl;
            cout<<"Send Kurdy"<<endl;
        }
        else
        {
            cout<<"Case #"<<k++<<':'<<endl;
            cout<<fixed<<setprecision(4)<<ans<<endl;
        }
        cout<<endl;
    }
    return 0;
}

Answer:

This question is very vague, the first example how to calculate out I have thought for a long while. In fact, the shortest path is calculated with floyed, and the longest of these shortest paths can be found.