图的顶点赋值问题

设G=（V，E）是有向无环图。写一个为G的顶点赋值的算法，如果存在一条从顶点i到顶点j的有向边，则i比j小

举例下图：

       |-------> b ------
    ---                        |
a                               ----------> d
    ---                        |
       |-------> c ------

有向无环图顶点 a b c d 的赋值应该为 12 3 4
====================================================================焦急等待算法结果的分割线====================================================================

#include <string>
#include <vector>
#include <iostream>

class vertex {
public:
    int head;
    int next;
    int in_degree;
    int val;

    vertex();
    vertex(int, int);
};

vertex::vertex()
{
    head = next = val = in_degree = 0;
}

vertex::vertex(int head, int next)
{
    this->head = head;
    this->next = next;
    in_degree = 0;
    val = 0;
}

std::vector<vertex> digraph;

void gen_digraph(std::vector<vertex>& dg) {
    dg.push_back(vertex(0, 0));
    dg.push_back(vertex(0, 5));
    dg.push_back(vertex(0, 7));
    dg.push_back(vertex(0, 0));
    dg.push_back(vertex(0, 8));
    dg.push_back(vertex(2, 6));
    dg.push_back(vertex(4, 0));
    dg.push_back(vertex(3, 0));
    dg.push_back(vertex(2, 9));
    dg.push_back(vertex(3, 0));

    std::vector<vertex>::iterator i;
    for (i = dg.begin(); i != dg.end(); i++) {
        if (i->head != 0) {
            int index = i->head;
            dg[index].in_degree++;
        }
    }
}

void topo_sort(std::vector<vertex>& dg, int vertex, int topo_val)
{
    int index = vertex;
    dg[index].val = topo_val;
    while (dg[index].next != 0) {
        int next = dg[index].next;
        int head = dg[next].head;
        dg[index].next = dg[next].next;
        if (--dg[head].in_degree == 0)
            topo_sort(dg, head, ++topo_val);
    }
}

int main(int argc, char** argv)
{
    int index = 1;
    std::vector<vertex>::iterator p;

    gen_digraph(digraph);

    for ( p = digraph.begin()+1; p->in_degree != 0 && p != digraph.end(); ++p, ++index);
    if (p == digraph.end())
        return -1;
    topo_sort(digraph, index, 1);

    index = 1;
    for ( p = digraph.begin()+1; p->head == 0 && p != digraph.end(); ++p, ++index)
        std::cout << index << "\t" << p->val << std::endl;

    return 0;
}