算法储备之有向无环图的最短路径问题

对于一般的加权图，Bellman-Ford计算单源点最短路径时间复杂度为O(VE),斐波那契堆优化后的Dijkstrs为O(E+VLogV)

如果图可以满足Directed Acyclic Graph(DAG)有向无环图，那么求单源点最短路径问题时间复杂度可以缩减到O(V+E)

同样这个方法可以解决多段图问题，多段图属于DAG。

初始化dist[]={INF,INF,...} and dist[src]=0;

对图进行拓扑排序

按拓扑排序的顺序对每一个顶点u执行

  if(dist[u]+weight(u,v)<dist[v])

    dist[v]=dist[u]+weight(u,v);

#include <iostream>#include <stack>#include <list>#include <vector>#include <functional>using namespace std;typedef pair<int, int> AdjListNode;class Graph {int V;list<AdjListNode> * adj;void topologicalSortUtil(int v, vector<bool>& visited, stack<int> & Stack);public:Graph(int _V) : V(_V),adj(new list<AdjListNode>[V]) { }void addEdge(int u, int v, int weight);void shortestPath(int s);};void Graph::addEdge(int u, int v, int weight){adj[u].push_back(AdjListNode(v, weight));}void Graph::topologicalSortUtil(int v,vector<bool>& visited, stack<int > & stack){visited[v] = true;list<AdjListNode>::iterator i;for (i = adj[v].begin();i != adj[v].end();i++)if (!visited[i->first])topologicalSortUtil(i->first, visited, stack);stack.push(v);}void printSolution(vector<int> & dist,int src){cout << "Following are shortest distances from source " << src << endl;for (int i = 0;i < dist.size();i++)(dist[i] == INT_MAX) ? cout << "INF " : cout << dist[i] << " ";cout << endl;}void Graph::shortestPath(int src){stack<int> stack;vector<int> dist(V, INT_MAX);vector<bool>visited(V, false);for (int i = 0;i < V;i++)if (visited[i] == false)topologicalSortUtil(i, visited, stack);dist[src] = 0;while (stack.empty() == false){int u = stack.top();stack.pop();list<AdjListNode>::iterator i;if (dist[u] != INT_MAX){for (i = adj[u].begin();i != adj[u].end();i++)if (dist[u] + i->second < dist[i->first])dist[i->first] = dist[u] + i->second;}}printSolution(dist, src);}int main(){Graph g(6);g.addEdge(0, 1, 5);g.addEdge(0, 2, 3);g.addEdge(1, 3, 6);g.addEdge(1, 2, 2);g.addEdge(2, 4, 4);g.addEdge(2, 5, 2);g.addEdge(2, 3, 7);g.addEdge(3, 4, -1);g.addEdge(4, 5, -2);g.shortestPath(2);return 0;}