The key to success is to focus our conscious mind on things we desire not things we fear.

Compute the minimum distance between all the nodes in a graph.

` ````
// n is the number of nodes in the graph, numbered from 0 to n - 1.
// Set D[i][j] = infinity for each (i, j).
// The distance of a node with itself is 0.
for (int i = 0; i < n; i++) {
D[i][i] = 0;
}
// W[i][j] is cost of the edge going from node i to j.
// Infinity if such edge doesn't exist.
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
D[i][j] = W[i][j];
}
}
for (int k = 0; k < n; k++) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
// Check if it is cheaper to travel from node i to node k
// and then from node k to node j, than traveling directly
// from i to j, update D if that's the case.
if (D[i][k] + D[k][j] < D[i][j]) {
D[i][j] = D[i][k] + D[k][j];
}
}
}
}
```

Quickly determine if node $i$ can reach node $j$ directly or by using some intermediate nodes, the answer is in $R[i][j]$.

` ````
// R is a copy of the adjacency matrix.
for (int k = 0; k < n; k++) {
R[k][k] = true;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (R[i][k] && R[k][j]) {
R[i][j] = true;
}
}
}
}
```

Algorithm complexity: $O(n^3)$

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