Algorithms/JavaAlgorithms
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

Improved readability

Browse Source
This commit is contained in:
Lucas Contini 2018-10-08 10:13:08 -03:00 committed by GitHub
parent 2c10f63f7a
commit 7ed62a9930
No known key found for this signature in database
GPG Key ID: 4AEE18F83AFDEB23
1 changed files with 69 additions and 47 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

116
Others/Dijkshtra.java
View File

@ -3,7 +3,6 @@
*/
import java.io.IOException;
import java.util.Arrays;
import java.util.Scanner;
@ -11,51 +10,74 @@ import java.util.Stack;
public class Dijkshtra {
public static void main(String[] args) throws IOException {
Scanner in =new Scanner(System.in);
int n=in.nextInt(); //n = Number of nodes or vertices
int m=in.nextInt(); //m = Number of Edges
long w[][]=new long [n+1][n+1]; //Adjacency Matrix
//Initializing Matrix with Certain Maximum Value for path b/w any two vertices
for (long[] row: w)
Arrays.fill(row, 1000000l);
//From above,we Have assumed that,initially path b/w any two Pair of vertices is Infinite such that Infinite = 1000000l
//For simplicity , We can also take path Value = Long.MAX_VALUE , but i have taken Max Value = 1000000l .
//Taking Input as Edge Location b/w a pair of vertices
for(int i=0;i<m;i++){
int x=in.nextInt(),y=in.nextInt();
long cmp=in.nextLong();
if(w[x][y]>cmp){ //Comparing previous edge value with current value - Cycle Case
w[x][y]=cmp; w[y][x]=cmp;
}
public static void main(String[] args) throws IOException {
Scanner in = new Scanner(System.in);
// n = Number of nodes or vertices
int n = in.nextInt();
// m = Number of Edges
int m = in.nextInt();
// Adjacency Matrix
long w[][] = new long [n+1][n+1];
//Initializing Matrix with Certain Maximum Value for path b/w any two vertices
for (long[] row : w) {
Arrays.fill(row, 1000000l);
}
/* From above,we Have assumed that,initially path b/w any two Pair of vertices is Infinite such that Infinite = 1000000l
For simplicity , We can also take path Value = Long.MAX_VALUE , but i have taken Max Value = 1000000l */
// Taking Input as Edge Location b/w a pair of vertices
for(int i = 0; i < m; i++) {
int x = in.nextInt(),y=in.nextInt();
long cmp = in.nextLong();
//Comparing previous edge value with current value - Cycle Case
if(w[x][y] > cmp) {
w[x][y] = cmp; w[y][x] = cmp;
}
}
// Implementing Dijkshtra's Algorithm
Stack<Integer> t = new Stack<Integer>();
int src = in.nextInt();
for(int i = 1; i <= n; i++) {
if(i != src) {
t.push(i);
}
}
Stack <Integer> p = new Stack<Integer>();
p.push(src);
w[src][src] = 0;
while(!t.isEmpty()) {
int min = 989997979;
int loc = -1;
for(int i = 0; i < t.size(); i++) {
w[src][t.elementAt(i)] = Math.min(w[src][t.elementAt(i)], w[src][p.peek()] + w[p.peek()][t.elementAt(i)]);
if(w[src][t.elementAt(i)] <= min) {
min = (int) w[src][t.elementAt(i)];
loc = i;
}
}
p.push(t.elementAt(loc));
t.removeElementAt(loc);
}
// Printing shortest path from the given source src
for(int i = 1; i <= n; i++) {
if(i != src && w[src][i] != 1000000l) {
System.out.print(w[src][i] + " ");
}
// Printing -1 if there is no path b/w given pair of edges
else if(i != src) {
System.out.print("-1" + " ");
}
}
}
//Implementing Dijkshtra's Algorithm
Stack<Integer> t=new Stack<Integer>();
int src=in.nextInt();
for(int i=1;i<=n;i++){
if(i!=src){t.push(i);}}
Stack <Integer> p=new Stack<Integer>();
p.push(src);
w[src][src]=0;
while(!t.isEmpty()){int min=989997979,loc=-1;
for(int i=0;i<t.size();i++){
w[src][t.elementAt(i)]=Math.min(w[src][t.elementAt(i)],w[src][p.peek()]
+w[p.peek()][t.elementAt(i)]);
if(w[src][t.elementAt(i)]<=min){
min=(int) w[src][t.elementAt(i)];loc=i;}
}
p.push(t.elementAt(loc));t.removeElementAt(loc);}
//Printing shortest path from the given source src
for(int i=1;i<=n;i++){
if(i!=src && w[src][i]!=1000000l){System.out.print(w[src][i]+" ");}
else if(i!=src){System.out.print("-1"+" ");} //Printing -1 if there is no path b/w given pair of edges
}
}
}