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====================================================================
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#include <string>
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#include <vector>
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#include <iostream>
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class vertex {
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public:
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int head;
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int next;
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int in_degree;
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int val;
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vertex();
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vertex(int, int);
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};
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vertex::vertex()
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{
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head = next = val = in_degree = 0;
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}
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vertex::vertex(int head, int next)
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{
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this->head = head;
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this->next = next;
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in_degree = 0;
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val = 0;
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}
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std::vector<vertex> digraph;
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void gen_digraph(std::vector<vertex>& dg) {
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dg.push_back(vertex(0, 0));
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dg.push_back(vertex(0, 5));
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dg.push_back(vertex(0, 7));
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dg.push_back(vertex(0, 0));
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dg.push_back(vertex(0, 8));
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dg.push_back(vertex(2, 6));
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dg.push_back(vertex(4, 0));
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dg.push_back(vertex(3, 0));
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dg.push_back(vertex(2, 9));
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dg.push_back(vertex(3, 0));
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std::vector<vertex>::iterator i;
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for (i = dg.begin(); i != dg.end(); i) {
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if (i->head != 0) {
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int index = i->head;
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dg[index].in_degree;
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}
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}
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}
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void topo_sort(std::vector<vertex>& dg, int vertex, int topo_val)
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{
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int index = vertex;
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dg[index].val = topo_val;
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while (dg[index].next != 0) {
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int next = dg[index].next;
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int head = dg[next].head;
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dg[index].next = dg[next].next;
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if (--dg[head].in_degree == 0)
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topo_sort(dg, head, topo_val);
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}
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}
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int main(int argc, char** argv)
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{
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int index = 1;
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std::vector<vertex>::iterator p;
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gen_digraph(digraph);
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for ( p = digraph.begin()1; p->in_degree != 0 && p != digraph.end(); p, index);
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if (p == digraph.end())
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return -1;
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topo_sort(digraph, index, 1);
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index = 1;
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for ( p = digraph.begin()1; p->head == 0 && p != digraph.end(); p, index)
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std::cout << index << "\t" << p->val << std::endl;
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return 0;
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}
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