Floyd Warshall Algorithm

int d[MAX_V][MAX_V];
for(int k = 0; k < MAX_V; ++k) {
    for(int i = 0; i < MAX_V; ++i) {
        for(int j = 0; j < MAX_V; ++j) {
            int newDist = d[i][k] + d[k][j];
            if (newDist < d[i][j]) {
                d[i][j] = newDist;
            }
        }
    }
}

Shortest Path Construction

int d[MAX_V][MAX_V];
int next[MAX_V][MAX_V];
for(int i = 0; i < MAX_V; ++i) {
    for(int j = 0; j < MAX_V; ++j) {
        next[i][j] = j;
    }
}
for(int k = 0; k < MAX_V; ++k) {
    for(int i = 0; i < MAX_V; ++i) {
        for(int j = 0; j < MAX_V; ++j) {
            int newDist = d[i][k] + d[k][j];
            if (newDist < d[i][j]) {
                d[i][j] = newDist;
                next[i][j] = next[i][k];
            }
        }
    }
}

vector<int> getPath(int i, int j) {
    vector<int> res; res.push_back(i);
    while(i != j) {
        res.push_back(next[i][j]);
        i = next[i][j];
    }
    return res;
}

Time Complexity

• Obviously, $O(V^3)$