It is a country with n cities. It is known that some cities have road connections. Ask which roads are added so that all cities can connect to each other at the lowest cost, the cost is the Cartesian distance between two city coordinates;

Is a pure question for finding the minimal spanning tree;

First, all edges are sorted by weight from small to large, and then the smallest edge is obtained in sequence. If the two nodes of China Unicom are on two Unicom components, add the edge; otherwise, delete the edge;

The key of the kruskal algorithm is to judge whether two nodes are in a single connected component, so there is enough flag array []; At the beginning, flag [I] = I; in this way, all nodes are connected by a single component, and then when the existing m edge is obtained,

Set the edge link node to u and v, and then set f [u] = f [v];. Then, the two nodes are connected to one connected component, however, the operation is not as simple as f [u] = f [v], because the UNICOM component of u must be connected with the UNICOM component of v, therefore, the flag value of each Unicom component can point to the same node. For specific operations, see the following code:

No order is required for the output edge of this question;

The reference code is as follows:

[Cpp]

# Include <iostream>

# Include <cstring>

# Include <algorithm>

# Include <cstdlib>

# Include <cmath>

Using namespace std;

Const int MAX_SIZE = 800;

Struct Node

{

Int u, v;

Double w;

Bool operator <(const Node a) const

{

Return w <a. w;

};

} Edge [MAX_SIZE * MAX_SIZE/2];

Struct point

{

Int x, y;

} A [MAX_SIZE];

Int flag [MAX_SIZE], num, n, m;

Int getflag (int u)

{

If (flag [u]! = U)

Flag [u] = getflag (flag [u]);

Return flag [u];

}

Void execute ()

{

Int I, j;

For (I = 1; I <= n; I ++)

{

For (j = I + 1; j <= n; j ++)

If (getflag (I )! = Getflag (j ))

{

Edge [num]. u = I;

Edge [num]. v = j;

Edge [num ++]. w = sqrt (a [I]. x-a [j]. x) * (a [I]. x-a [j]. x) + (a [I]. y-a [j]. y) * (a [I]. y-a [j]. y ));

}

}

Sort (edge, edge + num );

For (I = 0; I <num; I ++)

If (getflag (edge [I]. u )! = Getflag (edge [I]. v ))

{

Flag [getflag (edge [I]. u)] = flag [getflag (edge [I]. v)];

Cout <edge [I]. u <''<edge [I]. v <endl;

}

}

Int main ()

{

Cin> n;

Int u, v;

Int I;

For (I = 1; I <= n; I ++)

Cin> a [I]. x> a [I]. y;

Cin> m;

For (I = 1; I <= n + 2; I ++)

Flag [I] = I;

Num = 0;

While (m --)

{

Cin> u> v;

Flag [getflag (u)] = flag [getflag (v)];

}

Execute ();

Return 0;

}

# Include <iostream>

# Include <cstring>

# Include <algorithm>

# Include <cstdlib>

# Include <cmath>

Using namespace std;

Const int MAX_SIZE = 800;

Struct Node

{

Int u, v;

Double w;

Bool operator <(const Node a) const

{

Return w <a. w;

};

} Edge [MAX_SIZE * MAX_SIZE/2];

Struct point

{

Int x, y;

} A [MAX_SIZE];

Int flag [MAX_SIZE], num, n, m;

Int getflag (int u)

{

If (flag [u]! = U)

Flag [u] = getflag (flag [u]);

Return flag [u];

}

Void execute ()

{

Int I, j;

For (I = 1; I <= n; I ++)

{

For (j = I + 1; j <= n; j ++)

If (getflag (I )! = Getflag (j ))

{

Edge [num]. u = I;

Edge [num]. v = j;

Edge [num ++]. w = sqrt (a [I]. x-a [j]. x) * (a [I]. x-a [j]. x) + (a [I]. y-a [j]. y) * (a [I]. y-a [j]. y ));

}

}

Sort (edge, edge + num );

For (I = 0; I <num; I ++)

If (getflag (edge [I]. u )! = Getflag (edge [I]. v ))

{

Flag [getflag (edge [I]. u)] = flag [getflag (edge [I]. v)];

Cout <edge [I]. u <''<edge [I]. v <endl;

}

}

Int main ()

{

Cin> n;

Int u, v;

Int I;

For (I = 1; I <= n; I ++)

Cin> a [I]. x> a [I]. y;

Cin> m;

For (I = 1; I <= n + 2; I ++)

Flag [I] = I;

Num = 0;

While (m --)

{

Cin> u> v;

Flag [getflag (u)] = flag [getflag (v)];

}

Execute ();

Return 0;

}