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| 1 | +// C Program for Floyd Warshall Algorithm |
| 2 | +#include<stdio.h> |
| 3 | + |
| 4 | +// Number of vertices in the graph |
| 5 | +#define V 4 |
| 6 | + |
| 7 | +/* Define Infinite as a large enough value. This value will be used |
| 8 | + for vertices not connected to each other */ |
| 9 | +#define INF 99999 |
| 10 | + |
| 11 | +// A function to print the solution matrix |
| 12 | +void printSolution(int dist[][V]); |
| 13 | + |
| 14 | +// Solves the all-pairs shortest path problem using Floyd Warshall algorithm |
| 15 | +void floydWarshall (int graph[][V]) |
| 16 | +{ |
| 17 | + /* dist[][] will be the output matrix that will finally have the shortest |
| 18 | + distances between every pair of vertices */ |
| 19 | + int dist[V][V], i, j, k; |
| 20 | + |
| 21 | + /* Initialize the solution matrix same as input graph matrix. Or |
| 22 | + we can say the initial values of shortest distances are based |
| 23 | + on shortest paths considering no intermediate vertex. */ |
| 24 | + for (i = 0; i < V; i++) |
| 25 | + for (j = 0; j < V; j++) |
| 26 | + dist[i][j] = graph[i][j]; |
| 27 | + |
| 28 | + /* Add all vertices one by one to the set of intermediate vertices. |
| 29 | + ---> Before start of an iteration, we have shortest distances between all |
| 30 | + pairs of vertices such that the shortest distances consider only the |
| 31 | + vertices in set {0, 1, 2, .. k-1} as intermediate vertices. |
| 32 | + ----> After the end of an iteration, vertex no. k is added to the set of |
| 33 | + intermediate vertices and the set becomes {0, 1, 2, .. k} */ |
| 34 | + for (k = 0; k < V; k++) |
| 35 | + { |
| 36 | + // Pick all vertices as source one by one |
| 37 | + for (i = 0; i < V; i++) |
| 38 | + { |
| 39 | + // Pick all vertices as destination for the |
| 40 | + // above picked source |
| 41 | + for (j = 0; j < V; j++) |
| 42 | + { |
| 43 | + // If vertex k is on the shortest path from |
| 44 | + // i to j, then update the value of dist[i][j] |
| 45 | + if (dist[i][k] + dist[k][j] < dist[i][j]) |
| 46 | + dist[i][j] = dist[i][k] + dist[k][j]; |
| 47 | + } |
| 48 | + } |
| 49 | + } |
| 50 | + |
| 51 | + // Print the shortest distance matrix |
| 52 | + printSolution(dist); |
| 53 | +} |
| 54 | + |
| 55 | +/* A utility function to print solution */ |
| 56 | +void printSolution(int dist[][V]) |
| 57 | +{ |
| 58 | + printf ("The following matrix shows the shortest distances" |
| 59 | + " between every pair of vertices \n"); |
| 60 | + for (int i = 0; i < V; i++) |
| 61 | + { |
| 62 | + for (int j = 0; j < V; j++) |
| 63 | + { |
| 64 | + if (dist[i][j] == INF) |
| 65 | + printf("%7s", "INF"); |
| 66 | + else |
| 67 | + printf ("%7d", dist[i][j]); |
| 68 | + } |
| 69 | + printf("\n"); |
| 70 | + } |
| 71 | +} |
| 72 | + |
| 73 | +// driver program to test above function |
| 74 | +int main() |
| 75 | +{ |
| 76 | + /* Let us create the following weighted graph |
| 77 | + 10 |
| 78 | + (0)------->(3) |
| 79 | + | /|\ |
| 80 | + 5 | | |
| 81 | + | | 1 |
| 82 | + \|/ | |
| 83 | + (1)------->(2) |
| 84 | + 3 */ |
| 85 | + int graph[V][V] = { {0, 5, INF, 10}, |
| 86 | + {INF, 0, 3, INF}, |
| 87 | + {INF, INF, 0, 1}, |
| 88 | + {INF, INF, INF, 0} |
| 89 | + }; |
| 90 | + |
| 91 | + // Print the solution |
| 92 | + floydWarshall(graph); |
| 93 | + return 0; |
| 94 | +} |
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