Kruskal’s Minimum Spanning Tree Algorithm

Standard

Kruskal’s algorithm is the best and probably fastest option if you’re urging to form a subgraph of a graph connecting all nodes. Some implementations are shown.

Minimum Spanning Tree
To form an MST(Minimum Spanning Tree) follow this procedure,

1. Begin with a connected graph G.
2. E is the set of all edges of different weights, from G.
3. Sort edges of E in the ascending order of their weights.
4. Go through each member(edge) of E and see if the nodes of that edge is already connected or not.
5. If they aren’t connected, connect them and include that member of E(edge) int MST.
6. Continue this process untill you’ve got n-1 edges(in case of n nodes).

An implementation in C++ using Disjoint Set data structure, is shown

```#include <cstdio>
#include <algorithm>
#define MAX_node 100000
using namespace std;

struct edg{
int a,b,w;
bool operator < (const edg &b) const{
return w<b.w;
}
}E[MAX_node+1];

int Prev[MAX_node+1];
int parent(int a){
if(a==Prev[a]) return a;
return Prev[a]=parent(Prev[a]);
}

int main(){
int node,edge;
while(scanf("%d%d",&node,&edge)){

int TOTAL=0;

for(int i=1;i<=node;i++) Prev[i]=i;//*
for(int i=0;i<edge;i++) scanf("%d%d%d",&E[i].a,&E[i].b,&E[i].w);//**

sort(E,E+edge);//***

for(int i=0; i<edge; i++ ){
int u=parent(E[i].a);//****
int v=parent(E[i].b);

if(u!=v){
TOTAL+=E[i].w;
Prev[u]=v;//*****
}
}

printf("Total Cost of MST %d\n",TOTAL);
}
return 0;
}

/*
* Making each node it's own parent
** Input edges in the format "node1 node2 weight"
*** Sorting them in the ascending order of weight
**** Find the current parent of a and b
***** If the nodes of that edge is not connected yet, Connect them.
*/

```

if you don’t know Maximum possible number of nodes, then just use a vector(which is a bit slower) of type ‘edg’.

Maximum Spanning Tree
To get a maximum spanning tree, the procedure is almost same. The only difference is that, you got to sort members of E in the descending order of their weights.

Related Problems
Heavy Cargo(UVa)

Second Best Minimum Spanning Tree
There may be some other better procedure to find the second best MST, but this is the one I’ve used.
1. First form an MST of graph G, and mark the edges which formes the MST.
2. Then for each member edge of the MST, Form another MST without using that particular member.
3. Thus, u need to form n-1 (in case of n nodes) ‘another’ MSTs and consider the minimun among them.

Related Problems
Is There A Second Way Left ?(UVa)
ACM contest and Blackout(UVa)

Floyd Warshall

Standard

There are many notable algorithms to calculate the shortest path between vertices in a graph. Most are based on single source to a set of destination vertices. Examples of such famous algorithms include Dijkstra’s, Bellman-Ford and even Breadth First Search for weightless graphs. However, sometimes we wish to calculate the shortest paths between all pairs of vertices. An easy way to calculate this is to loop through each possible pair of vertices and run a Dijkstra through it. However, there exists a more efficient way to calculate this in n^3 time complexity.

Floyd-Warshall algorithm, a formula consisting of the core is very simple. A, B, C, I suppose we have 3 points. In this case, the shortest distance between A and C (distance (A, C), distance (A, B) + distance (B, C)) is up. Floyd-Warshall’ı need to apply to a table. The size of this table, depending on the number of points to be determined. We got one point, the table size N * N N

`int[,] table = new int[5,5];`

On the basis of the algorithm used for the function of the input for this table are the values at the maximum. The distance of each point of self-0, the distance to other points, we will begin by accepting the infinite.

```int INFINITY = 9999999;
int N = 5;
for (int i = 0; i < N; i++)
for(int j = 0; j < N; j++)
table[i,j] = i == j ? 0 : INFINITY;```

After this operation, the value of the table will be as follows.

 table 1 2 3 4 5 1 0 inf inf inf inf 2 inf 0 inf inf inf 3 inf inf 0 inf inf 4 inf inf inf 0 inf 5 inf inf inf inf 0

Second, to keep the connection between the points and to build the matrix of contiguity for questioning. If you have an edge from one point to another entry in the table that would give a value. For example, point 3 and point 4, one side has 2 points. Therefore, our statements [2,3] = [2.4] = 1 = [3.2] = [4.2]. Grafımız no way we obtain for the diagonal matrix based on contiguity should be symmetrical. 1 to 4, because to have the edge in the 4 to 1, this also means there are also edges. Our example is a way you could tell it had graphs. After this, the final version of the table will be as follows:

 table 1 2 3 4 5 1 0 inf inf 1 inf 2 inf 0 1 1 inf 3 inf 1 0 1 inf 4 1 1 1 0 1 5 inf inf inf 1 0

Floyd-Warshall can now apply the formula:

```for( int k = 0; k &lt; N; k++)
for( int i = 0; i &lt; N; i++)
for( int j = 0; j &lt; N; j++)
table[i,j] = Math.Min(table[i,j], table[i,k] + table[k,j]);```

That’s it. After this, our table will be as final.

 table 1 2 3 4 5 1 0 2 2 1 2 2 2 0 1 1 2 3 2 1 0 1 2 4 1 1 1 0 1 5 2 2 2 1 0

Now using the table between the two points (a, b) can query the shortest distance. table [a, b] = table [b, a].