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