Given a Weighted Directed Acyclic Graph and a source vertex in the graph, find the shortest paths from given source to all other vertices.
For a general weighted graph, we can calculate single source shortest distances in O(VE) time using Bellman–Ford Algorithm. For a graph with no negative weights, we can do better and calculate single source shortest distances in O(E + VLogV) time using Dijkstra’s algorithm. Can we do even better for Directed Acyclic Graph (DAG)? We can calculate single source shortest distances in O(V+E) time for DAGs. The idea is to use Topological Sorting.

We initialize distances to all vertices as infinite and distance to source as 0, then we find a topological sorting of the graph. Topological Sorting of a graph represents a linear ordering of the graph (See below, figure (b) is a linear representation of figure (a) ). Once we have topological order (or linear representation), we one by one process all vertices in topological order. For every vertex being processed, we update distances of its adjacent using distance of current vertex.

Following figure is taken from this source. It shows step by step process of finding shortest paths.

Following is complete algorithm for finding shortest distances.
1) Initialize dist[] = {INF, INF, ….} and dist[s] = 0 where s is the source vertex.
2) Create a toplogical order of all vertices.
3) Do following for every vertex u in topological order.
………..Do following for every adjacent vertex v of u
………………if (dist[v] > dist[u] + weight(u, v))
………………………dist[v] = dist[u] + weight(u, v)

// Java program to find single source shortest paths in Directed Acyclic Graphs
import java.io.*;
import java.util.*; class ShortestPath
{
static final int INF=Integer.MAX_VALUE;
class AdjListNode
{
private int v;
private int weight;
AdjListNode(int _v, int _w) { v = _v; weight = _w; }
int getV() { return v; }
int getWeight() { return weight; }
} // Class to represent graph as an adjcency list of
// nodes of type AdjListNode
class Graph
{
private int V;
private LinkedList<AdjListNode>adj[];
Graph(int v)
{
V=v;
adj = new LinkedList[V];
for (int i=0; i<v; ++i)
adj[i] = new LinkedList<AdjListNode>();
}
void addEdge(int u, int v, int weight)
{
AdjListNode node = new AdjListNode(v,weight);
adj[u].add(node);// Add v to u's list
} // A recursive function used by shortestPath.
// See below link for details
void topologicalSortUtil(int v, Boolean visited[], Stack stack)
{
// Mark the current node as visited.
visited[v] = true;
Integer i; // Recur for all the vertices adjacent to this vertex
Iterator<AdjListNode> it = adj[v].iterator();
while (it.hasNext())
{
AdjListNode node =it.next();
if (!visited[node.getV()])
topologicalSortUtil(node.getV(), visited, stack);
}
// Push current vertex to stack which stores result
stack.push(new Integer(v));
} // The function to find shortest paths from given vertex. It
// uses recursive topologicalSortUtil() to get topological
// sorting of given graph.
void shortestPath(int s)
{
Stack stack = new Stack();
int dist[] = new int[V]; // Mark all the vertices as not visited
Boolean visited[] = new Boolean[V];
for (int i = 0; i < V; i++)
visited[i] = false; // Call the recursive helper function to store Topological
// Sort starting from all vertices one by one
for (int i = 0; i < V; i++)
if (visited[i] == false)
topologicalSortUtil(i, visited, stack); // Initialize distances to all vertices as infinite and
// distance to source as 0
for (int i = 0; i < V; i++)
dist[i] = INF;
dist[s] = 0; // Process vertices in topological order
while (stack.empty() == false)
{
// Get the next vertex from topological order
int u = (int)stack.pop(); // Update distances of all adjacent vertices
Iterator<AdjListNode> it;
if (dist[u] != INF)
{
it = adj[u].iterator();
while (it.hasNext())
{
AdjListNode i= it.next();
if (dist[i.getV()] > dist[u] + i.getWeight())
dist[i.getV()] = dist[u] + i.getWeight();
}
}
} // Print the calculated shortest distances
for (int i = 0; i < V; i++)
{
if (dist[i] == INF)
System.out.print( "INF ");
else
System.out.print( dist[i] + " ");
}
}
} // Method to create a new graph instance through an object
// of ShortestPath class.
Graph newGraph(int number)
{
return new Graph(number);
} public static void main(String args[])
{
// Create a graph given in the above diagram. Here vertex
// numbers are 0, 1, 2, 3, 4, 5 with following mappings:
// 0=r, 1=s, 2=t, 3=x, 4=y, 5=z
ShortestPath t = new ShortestPath();
Graph g = t.newGraph(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); int s = 1;
System.out.println("Following are shortest distances "+
"from source " + s );
g.shortestPath(s);
}
}

Output:

Following are shortest distances from source 1
INF 0 2 6 5 3

Time Complexity: Time complexity of topological sorting is O(V+E). After finding topological order, the algorithm process all vertices and for every vertex, it runs a loop for all adjacent vertices. Total adjacent vertices in a graph is O(E). So the inner loop runs O(V+E) times. Therefore, overall time complexity of this algorithm is O(V+E).

algorithm@ Shortest Path in Directed Acyclic Graph (O(|V|+|E|) time)的更多相关文章

  1. 拓扑排序-有向无环图(DAG, Directed Acyclic Graph)

    条件: 1.每个顶点出现且只出现一次. 2.若存在一条从顶点 A 到顶点 B 的路径,那么在序列中顶点 A 出现在顶点 B 的前面. 有向无环图(DAG)才有拓扑排序,非DAG图没有拓扑排序一说. 一 ...

  2. AOJ GRL_1_C: All Pairs Shortest Path (Floyd-Warshall算法求任意两点间的最短路径)(Bellman-Ford算法判断负圈)

    题目链接:http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=GRL_1_C All Pairs Shortest Path Input ...

  3. AOJ GRL_1_A: Single Source Shortest Path (Dijktra算法求单源最短路径,邻接表)

    题目链接:http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=GRL_1_A Single Source Shortest Path In ...

  4. The Shortest Path in Nya Graph

    Problem Description This is a very easy problem, your task is just calculate el camino mas corto en ...

  5. (中等) HDU 4725 The Shortest Path in Nya Graph,Dijkstra+加点。

    Description This is a very easy problem, your task is just calculate el camino mas corto en un grafi ...

  6. HDU 4725 The Shortest Path in Nya Graph (最短路)

    The Shortest Path in Nya Graph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K ...

  7. Proof for Floyd-Warshall's Shortest Path Derivation Algorithm Also Demonstrates the Hierarchical Path Construction Process

    (THIS BLOG WAS ORIGINALLY WRTITTEN IN CHINESE WITH LINK: http://www.cnblogs.com/waytofall/p/3732920. ...

  8. The Shortest Path in Nya Graph HDU - 4725

    Problem Description This is a very easy problem, your task is just calculate el camino mas corto en ...

  9. HDU 4725 The Shortest Path in Nya Graph(最短路径)(2013 ACM/ICPC Asia Regional Online ―― Warmup2)

    Description This is a very easy problem, your task is just calculate el camino mas corto en un grafi ...

随机推荐

  1. Android title和actionbar的区别

    我想在一个页面的顶端放入两个按钮,应该用title还是actionbar.他们两个什么区别?分别该什么时候用? 答: android title 是UI上的一小部分,它支持Text和Color,你可以 ...

  2. 奇怪的transform bug

    对一个元素使用transform:rotate 进行旋转,造成: 父元素的背景图位置偏移,往下降,背景图也会变模糊一些 造成重绘,导致该元素后面的兄弟元素受到影响,变得模糊,并且无法遮盖住父元素的背景 ...

  3. 转:java提取图片中的像素

    本文转自:http://www.infosys.tuwien.ac.at/teaching/courses/WebEngineering/References/java/docs/api/java/a ...

  4. Codeforces Round #205 (Div. 2)

    A #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> ...

  5. 类handler

    /** The handler class is the interface for dynamically loadable storage engines. Do not add ifdefs a ...

  6. uva1639 Candy

    组合数,对数. 这道题要用到20w的组合数,如果直接相乘的话,会丢失很多精度,所以用去对数的方式实现. 注意指数,因为取完一次后,还要再取一次才能发现取完,所以是(n+1)次方. double 会爆掉 ...

  7. UVa 10088 (Pick定理) Trees on My Island

    这种1A的感觉真好 #include <cstdio> #include <vector> #include <cmath> using namespace std ...

  8. UVa 11609 (计数 公式推导) Teams

    n个人里选k个人有C(n, k)中方法,再从里面选一人当队长,有k中方法. 所以答案就是 第一步的变形只要按照组合数公式展开把n提出来即可. #include <cstdio> typed ...

  9. HDU 2433 Travel (最短路,BFS,变形)

    题意: 给出一个图的所有边,每次从图中删除一条边,求任意点对的路径总和(求完了就将边给补回去).(有重边) 思路: #include <bits/stdc++.h> using names ...

  10. 【C#学习笔记】播放wma/mp3文件

    using System; using System.Runtime.InteropServices; namespace ConsoleApplication { class Program { [ ...