Calculating critical path of a DAG in C++

I'm doing the calculation of the critical path for the DAG of the image, according to this algorithm for another post.My teacher requires that aarray be implemented, I simplify the homework statement, a simple graph implemented through arrays.

This es my code in which I have 3 arrays v, u and d, representing the origin node of the edges, the end node of the edges and the distance between each pair of vertices, as shown in the picture above. in the graph of the image, the duration of the project is equal to 25 corresponding to the sum of distances from the critical path.

My code fails to make good the calculation of distances according to the pseudocode of this link

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <queue>
#include <iostream>
#include <algorithm>

using namespace std;

int main (){


    int numberVertex=6; //number  of vertex
    int numberActivities=9;//number of edges
    int i, j;

    /*vertices are of the form (v, u) */
    //indegree of each vertex (that is, count the number of edges entering them) 
    int indegree [6]={0,0,0,0,0,0};

    int v[9]={0,0,1,1,2,4,3,3,3};//array v represent the starting vertex of de edge 
    int u[9]={1,2,2,3,4,5,4,5,5};//array u represent the coming vertex of de edge 
    int d[9]={5,6,3,8,2,12,0,1,4};//array d represent the time of de activity (v,u)
    int project_duration=0;//project duration


    /*# Compute the indegree for each vertex v from the graph:
    for each neighbor u of v: indegree[u] += 1*/
    for (j=0; j<numberActivities; j++){
        indegree[u[j]]++;
    }

    for (j=0;j<numberVertex; j++)
        printf ("indegree %d= %d\n",j,indegree[j] );

    queue<int> Q; //queue Q = empty queue
    int distance [numberVertex];
    memset(distance, 0, sizeof(int) * numberVertex);//distance = array filled with zeroes

    //for each vertex v:
    //if indegree[v] = 0:
    //insert v on Q
    for (j=0; j<numberVertex; j++)
    {
        if (indegree[j]==0)
            Q.push(v[j]);
    }

    int first;

    //printf ("first in the queue=%d\n", Q.front());

    /*for each neighbor u of v:
    d istance[u] = max(distance[u], distance[v] + time(v, u))
    indegree[u] -= 1
    if indegree[u] = 0:
    insert u on Q
    */
    while (!Q.empty()){ //while Q is not empty:
        first= Q.front ();  //v = get front element from Q
        Q.pop(); //delete de first from queue
        distance[u[first]]=std::max(distance[u[first]],
            distance[v[first]]+ d[first]);
        indegree[u[first]]-=1;

        if (indegree[u[first]]==0){

            Q.push(u[first]);
        }
    }



    for (j=0; j<numberVertex; j++)
    {
        printf ("dist [%d]= %d\n", j, distance[j]);
    }
    /*Now, select the vertex x with the largest distance. 
    This is the minimum total project_duration.*/
    printf ("Total Project Duration %d\n", project_duration);


    return (0);

}

What am I doing wrong or how it could solve the c开发者_StackOverflow中文版ode to tell me what is the duration of the project (corresponds to the sum of distances from the critical path)?. only able to calculate the distance to the first 3 vertex.

Your queue contains vertices. Your arrays u, v, d, are indexed by edge numbers. So you cannot write

first = Q.front();
... u[first] ...

since first is a vertex.

More generally, your code will be a lot easier to read (and the bug will be more obvious) if you use meaningful variable names. first is not very explicit (first what?), and u, v, d are also quite cryptic.

writing something like

cur_vertex = todo.front()
distance[dest[cur_vertex]] = std::max(distance[dest[cur_vertex]], 
    distance[source[cur_vertex]]+ weight[cur_vertex]);

will immediately raise a question: the source of a vertex, what is that? (Here we are using variable names as a substitute for proper type checking. An ADA programmer would have declared two different integer types to avoid the confusion between vertices and edge numbers.)

Another question: where did the loop over the successors of first go? It's in the pseudo-code, but not in your source.