|
| 1 | +// Floyd Warshall Algorithm |
| 2 | +// A space efficient algorithm to find shortest distance in space complexity of (O(1)) |
| 3 | +// Time complexity - O(v^3) |
| 4 | + |
| 5 | +#include<bits/stdc++.h> |
| 6 | +using namespace std; |
| 7 | + |
| 8 | +// Function which finds the shortest distance |
| 9 | +void shortest_distance(vector<vector<int>>&matrix) |
| 10 | +{ |
| 11 | + // Making first loop as intermediate between two vertices |
| 12 | + for(int k=0;k<matrix.size();k++) |
| 13 | + { |
| 14 | + // Value of i is treated as starting vertex of edge |
| 15 | + for(int i=0;i<matrix.size();i++) |
| 16 | + { |
| 17 | + // Value of j is treated as starting vertex of edge |
| 18 | + for(int j=0;j<matrix.size();j++) |
| 19 | + { |
| 20 | + // Avoiding self edge as well as intermediate vertex which is same as either of the two vertices of edge |
| 21 | + if(k!=i && k!=j && i!=j) |
| 22 | + { |
| 23 | + // Not trying to put unavailable edge as intermediate edge And also adding edge if not available |
| 24 | + if(matrix[i][k]!=-1 && matrix[k][j]!=-1 && ((matrix[i][j] > (matrix[i][k]+matrix[k][j])) || matrix[i][j]==-1)) |
| 25 | + matrix[i][j]=matrix[i][k]+matrix[k][j]; |
| 26 | + } |
| 27 | + } |
| 28 | + |
| 29 | + } |
| 30 | + } |
| 31 | +} |
| 32 | + |
| 33 | +int main() |
| 34 | +{ |
| 35 | + int v; |
| 36 | + cin>>v; |
| 37 | + vector<vector<int>>matrix(v,vector<int>(v,-1)); |
| 38 | + for(int i=0;i<v;i++) |
| 39 | + { |
| 40 | + for(int j=0;j<v;j++) |
| 41 | + { |
| 42 | + cin>>matrix[i][j]; |
| 43 | + } |
| 44 | + } |
| 45 | + shortest_distance(matrix); |
| 46 | + |
| 47 | + // Printing the updated matrix |
| 48 | + for(int i=0;i<v;i++) |
| 49 | + { |
| 50 | + for(int j=0;j<v;j++) |
| 51 | + { |
| 52 | + cout<<matrix[i][j]<<" "; |
| 53 | + } |
| 54 | + cout<<endl; |
| 55 | + } |
| 56 | + return 0; |
| 57 | +} |
| 58 | + |
| 59 | +// Description |
| 60 | +// This algorithm helps in finding shortest distance between |
| 61 | +// every pair of vertices in a given edge weighted directed Graph in in-place. |
| 62 | +// Core idea:- add any other vertex in between two edge and check if the previous distance |
| 63 | +// is greater than adding two edge connected with the intermediate edge, if yes update it. |
0 commit comments