GYM 100827 L.Wormhole（Floyd）

Description

给出n个星球的名称和三维坐标，距离定义为欧氏距离，其中有m对星球之间存在虫洞，那么这对星球之间距离为0，q次查询，每次查询一对星球之间的最短距离

Input

第一行一整数T表示用例组数，每组用例首先输入一整数n表示星球数量，之后输入每个星球的名字和三维坐标x,y,x，然后是一整数m表示虫洞数量，之后m行每行输入两个星球的名字表示这对星球之间存在虫洞，然后是一整数q表示查询数，之后q行每行两个星球的名字表示查询这两个星球的最短距离(1<=T<=10,1<=n<=60,0<=x,y,z<=2e6,0<=m<=40,1<=q<=20,名字不超过50个字符)

Output

对于每次查询，输出这两个星球之间最短距离

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个点，建好图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;
}