Description
給出n個星球的名稱和三維座標，距離定義為歐氏距離，其中有m對星球之間存在蟲洞，那麼這對星球之間距離為0，q次查詢，每次查詢一對星球之間的最短距離
Input
第一行一整數T表示用例組數，每組用例首先輸入一整數n表示星球數量，之後輸入每個星球的名字和三維座標x,y,x，然後是一整數m表示蟲洞數量，之後m行每行輸入兩個星球的名字表示這對星球之間存在蟲洞，然後是一整數q表示查詢數，之後q行每行兩個星球的名字表示查詢這兩個星球的最短距離(1<=T<=10,1<=n<=60,0<=x,y,z<=2e6,0<=m<=40,1<=q<=20,名字不超過50個字元)
Output

對於每次查詢，輸出這兩個星球之間最短距離
Sample Input
3
4
Earth 0 0 0
Proxima 5 0 0
Barnards 5 5 0
Sirius 0 5 0
2
Earth Barnards
Barnards Sirius
6
Earth Proxima
Earth Barnards
Earth Sirius
Proxima Earth
Barnards Earth
Sirius Earth
3
z1 0 0 0
z2 10 10 10
z3 10 0 0
1
z1 z2
3
z2 z1
z1 z2
z1 z3
2
Mars 12345 98765 87654
Jupiter 45678 65432 11111
0
1
Mars Jupiter
Sample Output
Case 1:
The distance from Earth to Proxima is 5 parsecs.
The distance from Earth to Barnards is 0 parsecs.
The distance from Earth to Sirius is 0 parsecs.
The distance from Proxima to Earth is 5 parsecs.
The distance from Barnards to Earth is 5 parsecs.
The distance from Sirius to Earth is 5 parsecs.
Case 2:
The distance from z2 to z1 is 17 parsecs.
The distance from z1 to z2 is 0 parsecs.
The distance from z1 to z3 is 10 parsecs.
Case 3:
The distance from Mars to Jupiter is 89894 parsecs.
Solution
才60個點，建好圖Floyd跑一遍即可
Code

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<ctime>
using namespace std;
typedef long long ll;
#define INF 0x3f3f3f3f
#define maxn 66
map<string,int>M;
int T,n,m,q,x[maxn],y[maxn],z[maxn];
double dis[maxn][maxn];
string s[maxn];
double get(int i,int j)
{
    return sqrt(pow(x[i]-x[j],2)+pow(y[i]-y[j],2)+pow(z[i]-z[j],2));
}
int main()
{
    scanf("%d",&T);
    for(int Case=1;Case<=T;Case++)
    {
        M.clear();
        scanf("%d",&n);
        for(int i=1;i<=n;i++) 
        {
            cin>>s[i]>>x[i]>>y[i]>>z[i];
            M[s[i]]=i;
        }
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                dis[i][j]=get(i,j);
        scanf("%d",&m);
        while(m--)
        {
            string a,b;
            cin>>a>>b;
            dis[M[a]][M[b]]=0;
        }
        for(int k=1;k<=n;k++)
            for(int i=1;i<=n;i++)
                for(int j=1;j<=n;j++)
                    dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
        scanf("%d",&q);
        printf("Case %d:\n",Case);
        while(q--)
        {
            string a,b;
            char c[maxn],d[maxn];
            scanf("%s%s",c,d);
            a=c,b=d;
            printf("The distance from %s to %s is %.0f parsecs.\n",c,d,dis[M[a]][M[b]]);
        }
    }
    return 0;
}